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# 3.8: Inverses and Radical Functions

In this section, you will:
• Find the inverse of a polynomial function.
• Restrict the domain to find the inverse of a polynomial function.

A mound of gravel is in the shape of a cone with the height equal to twice the radius.

<figure class="small" id="Figure_03_08_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"></figure>

The volume is found using a formula from elementary geometry.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>V</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 3 π r 2 h    = 1 3 π r 2 (2r)    = 2 3 π r 3

We have written the volume<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of the radius<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mroot/></mrow></annotation-xml></semantics>[/itex] 3V 2π 3

This function is the inverse of the formula for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.

# Finding the Inverse of a Polynomial Function

Two functions<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>g</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]are inverse functions if for every coordinate pair in<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo><mo> </mo><mo stretchy="false">(</mo><mi>a</mi><mo>,</mo><mo> </mo><mi>b</mi><mo stretchy="false">)</mo><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]there exists a corresponding coordinate pair in the inverse function,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>g</mi><mo>,</mo><mo stretchy="false">(</mo><mi>b</mi><mo>,</mo><mtext> </mtext><mi>a</mi><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]In other words, the coordinate pairs of the inverse functions have the input and output interchanged.

For a function to have an inverse function the function to create a new function that is one-to-one and would have an inverse function.

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in [link]. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

<figure class="small" id="Figure_03_08_002"></figure>

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]measured horizontally and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]measured vertically, with the origin at the vertex of the parabola. See[link].

<figure class="small" id="Figure_03_08_003"></figure>

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics>[/itex] x 2 . Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>18</mn><mo>=</mo><mi>a</mi><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 6 2  a= 18 36      = 1 2

Our parabolic cross section has the equation

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 2 x 2

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the width will be given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>x</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]so we need to solve the equation above for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]values. On this domain, we can find an inverse by solving for the input variable:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 1 2 x 2 2y= x 2   x=± 2y

This is not a function as written. We are limiting ourselves to positive<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]values, so we eliminate the negative solution, giving us the inverse function we’re looking for.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 2 , x>0

Because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the distance from the center of the parabola to either side, the entire width of the water at the top will be<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The trough is 3 feet (36 inches) long, so the surface area will then be:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>Area</mtext><mo>=</mo><mi>l</mi><mo>⋅</mo><mi>w</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]         =36⋅2x         =72x         =72 2y

This example illustrates two important points:

1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.

Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x).

Warning:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x) is not the same as the reciprocal of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ). This use of “–1” is reserved to denote inverse functions. To denote the reciprocal of a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ), we would need to write<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] ( f( x ) ) −1 = 1 f( x ) .

An important relationship between inverse functions is that they “undo” each other. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1  is the inverse of a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 . In other words, whatever the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]does to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1  undoes it—and vice-versa. More formally, we write

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 ( f( x ) )=x, for all x in the domain of f

and

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] f −1 ( x ) )=x, for all x in the domain of  f −1
Verifying Two Functions Are Inverses of One Another

Two functions,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>g</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]are inverses of one another if for all<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in the domain of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>g</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>g</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] f( x ) )=f( g( x ) )=x

Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.

1. Replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
2. Interchange<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
3. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and rename the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x).
Verifying Inverse Functions

Show that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 1 x+1  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 ( x )= 1 x −1 are inverses, for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≠</mo><mn>0</mn><mo>,</mo><mo>−</mo><mn>1</mn></mrow></annotation-xml></semantics>[/itex].

We must show that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 ( f( x ) )=x and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] f −1 ( x ) )=x.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>f</mi></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −1 (f(x))= f −1 ( 1 x+1 )                 = 1 1 x+1 −1                 =(x+1)−1                 =x f( f −1 (x))=f( 1 x −1 )                 = 1 ( 1 x−1 )+1                 = 1 1 x                 =x

Therefore,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 1 x+1   and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 ( x )= 1 x −1 are inverses.

Show that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x+5 3   and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 ( x )=3x−5 are inverses.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 ( f( x ) )= f −1 ( x+5 3 )=3( x+5 3 )−5=( x−5 )+5=x and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] f −1 ( x ) )=f( 3x−5 )= ( 3x−5 )+5 3 = 3x 3 =x

Finding the Inverse of a Cubic Function

Find the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>5</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +1.

This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>         </mtext><mi>y</mi><mo>=</mo><mn>5</mn><msup/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 3 +1          x=5 y 3 +1    x−1=5 y 3    x−1 5 = y 3 f −1 (x)= x−1 5 3
Analysis

Look at the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f –1 . Notice that the two graphs are symmetrical about the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]This is always the case when graphing a function and its inverse function.

Also, since the method involved interchanging<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]notice corresponding points. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>b</mi><mo>,</mo><mi>a</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f –1 . Since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mn>1</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>1</mn><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f –1 . Similarly, since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>1</mn><mo>,</mo><mn>6</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>6</mn><mo>,</mo><mn>1</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f –1 . See [link].

<figure class="small" id="Figure_03_08_004"></figure>

Find the inverse function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mroot/></mrow></annotation-xml></semantics>[/itex] x+4 3 .

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x 3 −4

# Restricting the Domain to Find the Inverse of a Polynomial Function

So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have aninverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.

Restricting the Domain

If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.

Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

1. Restrict the domain by determining a domain on which the original function is one-to-one.
2. Replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo> </mo><mtext>with</mtext><mo> </mo><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
3. Interchange<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo> </mo><mtext>and</mtext><mo> </mo><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
4. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and rename the function or pair of function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x).
5. Revise the formula for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.
Restricting the Domain to Find the Inverse of a Polynomial Function

Find the inverse function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>:</mo></mrow></annotation-xml></semantics>[/itex]

1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] (x−4) 2 , x≥4
2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] (x−4) 2 , x≤4

The original function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] (x−4) 2  is not one-to-one, but the function is restricted to a domain of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>4</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≤</mo><mn>4</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]on which it is one-to-one. See [link].

<figure class="medium" id="Figure_03_08_005"></figure>

To find the inverse, start by replacing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with the simple variable<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>y</mi><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (x−4) 2 Interchange x and y.              x= (y−4) 2 Take the square root.     ± x =y−4  Add 4 to both sides. 4± x =y

This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the original<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we looked at the domain: the values<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] could assume. When we reversed the roles of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] this gave us the values<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]could assume. For this function,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>4</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]so for the inverse, we should have<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>≥</mo><mn>4</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]which is what our inverse function gives.

1. The domain of the original function was restricted to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>4</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]so the outputs of the inverse need to be the same,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )≥4, and we must use the + case:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)=4+ x
2. The domain of the original function was restricted to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≤</mo><mn>4</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]so the outputs of the inverse need to be the same,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )≤4, and we must use the – case:
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)=4− x
Analysis

On the graphs in [link], we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The coordinate pair<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>4</mn><mo>,</mo><mo> </mo><mn>0</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and the coordinate pair<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mo> </mo><mn>4</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 . For any coordinate pair, if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a, b ) is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] b, a ) is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 . Finally, observe that the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]intersects the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1on the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Points of intersection for the graphs of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1  will always lie on the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<figure id="Figure_03_08_006"></figure>
Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified

Restrict the domain and then find the inverse of

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] (x−2) 2 −3.

We can see this is a parabola with vertex at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>2</mn><mo>,</mo><mo> </mo><mo>–</mo><mn>3</mn><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>2.</mn></mrow></annotation-xml></semantics>[/itex]

To find the inverse, we will use the vertex form of the quadratic. We start by replacing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with a simple variable,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>y</mi><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (x−2) 2 −3 Interchange x and y.                          x= (y−2) 2 −3 Add 3 to both sides.                x+3= (y−2) 2Take the square root.    ± x+3 =y−2 Add 2 to both sides. 2± x+3 =y Rename the function.            f −1 (x)=2± x+3

Now we need to determine which case to use. Because we restricted our original function to a domain of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>2</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the outputs of the inverse should be the same, telling us to utilize the + case

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)=2+ x+3

If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

Analysis

Notice that we arbitrarily decided to restrict the domain on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]We could just have easily opted to restrict the domain on<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≤</mo><mn>2</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in which case<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)=2− x+3 . Observe the original function graphed on the same set of axes as its inverse function in[link]. Notice that both graphs show symmetry about the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The coordinate pair<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2, −3 ) is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and the coordinate pair<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3, 2 ) is on the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 . Observe from the graph of both functions on the same set of axes that

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>domain of </mtext><mi>f</mi><mo>=</mo><mtext>range of</mtext><mo> </mo><msup/></mrow></annotation-xml></semantics>[/itex] f –1 =[ 2,∞ )

and

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>domain of </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f –1 =range of f=[ –3,∞ )

Finally, observe that the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]intersects the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1  along the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<figure class="small" id="Figure_03_08_007"></figure>

Find the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +1, on the domain<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>0.</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x−1

## Solving Applications of Radical Functions

Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited.

Given a radical function, find the inverse.

1. Determine the range of the original function.
2. Replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )  with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]
3. If necessary, restrict the domain of the inverse function to the range of the original function.
Finding the Inverse of a Radical Function

Restrict the domain and then find the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] x−4 .

Note that the original function has range<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>≥</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr columnalign="left"><mtd columnalign="right"><mi>y</mi></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] = x−4 Replace f(x) with y. x = y−4 Interchange x and y. x = y−4 Square each side. x 2 =y−4 Add 4. x 2 +4 =y Rename the function f −1 (x). f −1 (x) = x 2 +4

Recall that the domain of this function must be limited to the range of the original function.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x 2 +4,x≥0
Analysis

Notice in [link] that the inverse is a reflection of the original function over the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Because the original function has only positive outputs, the inverse function has only positive inputs.

<figure class="small" id="Figure_03_08_008"></figure>

Restrict the domain and then find the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 2x+3 .

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x 2 −3 2 ,x≥0

## Solving Applications of Radical Functions

Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.

Solving an Application with a Cubic Function

A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>V</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 3 π r 3

Find the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 3 π r 3  that determines the volume<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of a cone and is a function of the radius<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>π</mi><mo>=</mo><mn>3.14.</mn></mrow></annotation-xml></semantics>[/itex]

Start with the given function for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Notice that the meaningful domain for the function is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>≥</mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>≥</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using the method outlined previously.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>V</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 3 π r 3 r 3 = 3V 2π Solve for  r 3 .    r= 3V 2π 3 Solve for r.

This is the result stated in the section opener. Now evaluate this for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>=</mo><mn>100</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>π</mi><mo>=</mo><mn>3.14.</mn></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>r</mi><mo>=</mo><mroot/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3V 2π 3           = 3⋅100 2⋅3.14 3     ≈ 47.7707 3   ≈3.63

## Determining the Domain of a Radical Function Composed with Other Functions

When radical functions are composed with other functions, determining domain can become more complicated.

Finding the Domain of a Radical Function Composed with a Rational Function

Find the domain of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] (x+2)(x−3) (x−1) .

Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] (x+2)(x−3) (x−1) ≥0. The output of a rational function can change signs (change from positive to negative or vice versa) atx-intercepts and at vertical asymptotes. For this equation, the graph could change signs at <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi></mrow></annotation-xml></semantics>[/itex] = –2, 1, and 3.

To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in [link].

<figure class="medium" id="Figure_03_08_009"></figure>

This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a y-intercept at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mo> </mo><mn>6</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

From the y-intercept and x-intercept at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mn>2</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.

From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) will be defined.<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) has domain<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mn>2</mn><mo>≤</mo><mi>x</mi><mo><</mo><mn>1</mn><mtext> </mtext><mtext>or</mtext><mtext> </mtext><mi>x</mi><mo>≥</mo><mn>3</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or in interval notation,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">[</mo><mo>−</mo><mn>2</mn><mo>,</mo><mn>1</mn><mo stretchy="false">)</mo><mo>∪</mo><mo stretchy="false">[</mo><mn>3</mn><mo>,</mo><mi>∞</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

## Finding Inverses of Rational Functions

As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.

Finding the Inverse of a Rational Function

The function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 20+0.4n 100+n  represents the concentration<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of an acid solution after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.

We first want the inverse of the function. We will solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>C</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 20+0.4n 100+n   C(100+n)=20+0.4n   100C+Cn=20+0.4n       100C−20=0.4n−Cn        100C−20=(0.4−C)n                               n= 100C−20 0.4−C

Now evaluate this function for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>=</mo><mn>0.35</mn><mo> </mo><mo stretchy="false">(</mo><mn>35</mn><mi>%</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>n</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 100(0.35)−20 0.4−0.35    = 15 0.05 =300

We can conclude that 300 mL of the 40% solution should be added.

Find the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] x+3 x−2 .

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= 2x+3 x−1

Access these online resources for additional instruction and practice with inverses and radical functions.

# Key Concepts

• The inverse of a quadratic function is a square root function.
• If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1   is the inverse of a function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] is the inverse of the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 .  See [link].
• While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See[link].
• To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See[link] and [link].
• Inverse and radical and functions can be used to solve application problems. See [link] and [link].

# Section Exercises

## Verbal

Explain why we cannot find inverse functions for all polynomial functions.

It can be too difficult or impossible to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Why must we restrict the domain of a quadratic function when finding its inverse?

When finding the inverse of a radical function, what restriction will we need to make?

We will need a restriction on the domain of the answer.

The inverse of a quadratic function will always take what form?

## Algebraic

For the following exercises, find the inverse of the function on the given domain.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= ( x−4 ) 2 , [4,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= x +4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= ( x+2 ) 2 , [−2,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] ( x+1 ) 2 −3, [−1,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= x+3 −1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mo>−</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 3+x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>3</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +5,  ( − ∞,0 ],[0,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)=− x−5 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=12− x 2 , [0,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=9− x 2 , [0,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 9−x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +4, [0,∞)

For the following exercises, find the inverse of the functions.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +5

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= x−5 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=3 x 3 +1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mo>−</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= 4−x 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=4−2 x 3

For the following exercises, find the inverse of the functions.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 2x+1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x 2 −1 2 ,  [ 0,∞ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 3−4x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=9+ 4x−4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= ( x−9 ) 2 +4 4 ,  [ 9,∞ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 6x−8 +5

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=9+2 x 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= ( x−9 2 ) 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=3− x 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 2 x+8

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= 2−8x x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 3 x−4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x+3 x+7

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= 7x−3 1−x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= x−2 x+7

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 3x+4 5−4x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] f −1 (x)= 5x−4 4x+3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )= 5x+1 2−5x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +2x, [−1,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x+1 −1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +4x+1, [−2,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −6x+3, [3,∞)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x+6 +3

## Graphical

For the following exercises, find the inverse of the function and graph both the function and its inverse.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +2, x≥0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mo>−</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 , x≥0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= 4−x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] ( x+3 ) 2 , x≥−3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] ( x−4 ) 2 , x≥4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x +4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mo>−</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= 1−x 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +4x, x≥−2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −6x+1, x≥3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x+8 +3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 x 2 , x≥0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= 1 x

For the following exercises, use a graph to help determine the domain of the functions.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] (x+1)(x−1) x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] (x+2)(x−3) x−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">[</mo><mo>−</mo><mn>2</mn><mo>,</mo><mn>1</mn><mo stretchy="false">)</mo><mo>∪</mo><mo stretchy="false">[</mo><mn>3</mn><mo>,</mo><mi>∞</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] x(x+3) x−4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] x 2 −x−20 x−2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">[</mo><mo>−</mo><mn>4</mn><mo>,</mo><mn>2</mn><mo stretchy="false">)</mo><mo>∪</mo><mo stretchy="false">[</mo><mn>5</mn><mo>,</mo><mi>∞</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 9− x 2 x+4

## Technology

For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with y-coordinates given.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3 −x−2, y=1, 2, 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mo>–</mo><mn>2</mn><mo>,</mo><mo> </mo><mn>0</mn><mo stretchy="false">)</mo><mo>;</mo><mo> </mo><mo stretchy="false">(</mo><mn>4</mn><mo>,</mo><mo> </mo><mn>2</mn><mo stretchy="false">)</mo><mo>;</mo><mo> </mo><mo stretchy="false">(</mo><mn>22</mn><mo>,</mo><mo> </mo><mn>3</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +x−2, y=0, 1, 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +3x−4, y=0, 1, 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mo>–</mo><mn>4</mn><mo>,</mo><mo> </mo><mn>0</mn><mo stretchy="false">)</mo><mo>;</mo><mo> </mo><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mo> </mo><mn>1</mn><mo stretchy="false">)</mo><mo>;</mo><mo> </mo><mo stretchy="false">(</mo><mn>10</mn><mo>,</mo><mo> </mo><mn>2</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +8x−4, y=−1, 0, 1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 4 +5x+1, y=−1, 0, 1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mo>–</mo><mn>3</mn><mo>,</mo><mo> </mo><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>;</mo><mo> </mo><mo stretchy="false">(</mo><mn>1</mn><mo>,</mo><mo> </mo><mn>0</mn><mo stretchy="false">)</mo><mo>;</mo><mo> </mo><mo stretchy="false">(</mo><mn>7</mn><mo>,</mo><mo> </mo><mn>1</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]

## Extensions

For the following exercises, find the inverse of the functions with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>,</mo><mi>b</mi><mo>,</mo><mi>c</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]positive real numbers.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics>[/itex] x 3 +b

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +bx

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x+ b 2 4 − b 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a x 2 +b

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mroot/></mrow></annotation-xml></semantics>[/itex] ax+b 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>f</mi></msup></mrow></annotation-xml></semantics>[/itex] −1 (x)= x 3 −b a

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] ax+b x+c

## Real-World Applications

For the following exercises, determine the function described and then use it to answer the question.

An object dropped from a height of 200 meters has a height,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] t ), in meters after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]seconds have lapsed, such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>200</mn><mo>−</mo><mn>4.9</mn><msup/></mrow></annotation-xml></semantics>[/itex] t 2 . Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of height,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the time to reach a height of 50 meters.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo stretchy="false">(</mo><mi>h</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 200−h 4.9 , 5.53 seconds

An object dropped from a height of 600 feet has a height,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] t ), in feet after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]seconds have elapsed, such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>600</mn><mo>−</mo><mn>16</mn><msup/></mrow></annotation-xml></semantics>[/itex] t 2 . Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex] as a function of height<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the time to reach a height of 400 feet.

The volume,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of a sphere in terms of its radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo stretchy="false">(</mo><mi>r</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 4 3 π r 3 . Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the radius of a sphere with volume of 200 cubic feet.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo stretchy="false">(</mo><mi>V</mi><mo stretchy="false">)</mo><mo>=</mo><mroot/></mrow></annotation-xml></semantics>[/itex] 3V 4π 3 , 3.63 feet

The surface area,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of a sphere in terms of its radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo stretchy="false">(</mo><mi>r</mi><mo stretchy="false">)</mo><mo>=</mo><mn>4</mn><mi>π</mi><msup/></mrow></annotation-xml></semantics>[/itex] r 2 . Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the radius of a sphere with a surface area of 1000 square inches.

A container holds 100 ml of a solution that is 25 ml acid. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]ml of a solution that is 60% acid is added, the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 25+.6n 100+n  gives the concentration,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of the number of ml added,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and determine the number of mL that need to be added to have a solution that is 50% acid.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>n</mi><mo stretchy="false">(</mo><mi>C</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 100C−25 .6−C , 250 mL

The period<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in seconds, of a simple pendulum as a function of its length<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>l</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in feet, is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo stretchy="false">(</mo><mi>l</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mi>π</mi><msqrt/></mrow></annotation-xml></semantics>[/itex] l 32.2   . Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>l</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and determine the length of a pendulum with period of 2 seconds.

The volume of a cylinder ,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and height,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>=</mo><mi>π</mi><msup/></mrow></annotation-xml></semantics>[/itex] r 2 h. If a cylinder has a height of 6 meters, express the radius as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the radius of a cylinder with volume of 300 cubic meters.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo stretchy="false">(</mo><mi>V</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] V 6π , 3.99 meters

The surface area,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]of a cylinder in terms of its radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and height,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mn>2</mn><mi>π</mi><msup/></mrow></annotation-xml></semantics>[/itex] r 2 +2πrh. If the height of the cylinder is 4 feet, express the radius as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the radius if the surface area is 200 square feet.

The volume of a right circular cone,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of its radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and its height,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 3 π r 2 h. Express<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]if the height of the cone is 12 feet and find the radius of a cone with volume of 50 cubic inches.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo stretchy="false">(</mo><mi>V</mi><mo stretchy="false">)</mo><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] V 4π , 1.99 inches

Consider a cone with height of 30 feet. Express the radius,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in terms of the volume,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>V</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and find the radius of a cone with volume of 1000 cubic feet.

## Glossary

invertible function
any function that has an inverse function