2.8: Revisiting Inverses
In this section, you will:
- Restrict the domain to find the inverse of a polynomial function.
- Find the inverse of a simple rational function.
A mound of gravel is in the shape of a cone with the height equal to twice the radius.
Figure \( \PageIndex{ 1 } \)
The volume is found using a formula from Geometry.\[ \begin{array}{rcl} V & = & \dfrac{1}{3} \pi r^2 h \\[6pt] & = & \dfrac{1}{3} \pi r^2 (2r) \\[6pt] & = & \dfrac{2}{3} \pi r^3 \\[6pt] \end{array} \nonumber \]We have written the volume \(V\) in terms of the radius \(r\). 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\[ r = \sqrt[3]{ \dfrac{3V}{2 \pi}} \nonumber \]This function is the inverse of the formula for \(V\) in terms of \(r\).
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 \(f\) and \(g\) are inverse functions if for every coordinate pair in \(f\), \((a,b)\), there exists a corresponding coordinate pair in the inverse function, \(g\), \((b,a)\). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.
For a function to have an inverse, it must be one-to-one.
For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure \( \PageIndex{ 2 } \). 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 \( \PageIndex{ 2 } \)
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 \(x\) measured horizontally and \(y\) measured vertically, with the origin at the vertex of the parabola. See Figure \( \PageIndex{ 3 } \).
Figure \( \PageIndex{ 3 } \)
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 \(y(x)= a x^2 \). Our equation will need to pass through the point \((6, 18)\), from which we can solve for the stretch factor \(a\).\[ \begin{array}{rrcl} & 18 & = & a 6^2 \\[6pt] \implies & 18 & = & 36a \\[6pt] \implies & \dfrac{1}{2} & = & a \\[6pt] \end{array} \nonumber \]Our parabolic cross section has the equation\[y(x)= \dfrac{1}{2} x^2. \nonumber \]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 \(y\) the width will be given by \(2x\), so we need to solve the equation above for \(x\) 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 \(x\) values. On this domain, we can find an inverse by solving for the input variable:\[ \begin{array}{rrcl} & y & = & \dfrac{1}{2} x^2 \\[6pt] \implies & 2y & = & x^2 \\[6pt] \implies & \pm \sqrt{2y} & = & x \\[6pt] \end{array} \nonumber \]This is not a function as written. We are limiting ourselves to positive \(x\) values, so we eliminate the negative solution, giving us the inverse function we're looking for.\[y = \sqrt{2 x}, \text{ where } x > 0. \nonumber \]Because \(x\) is the distance from the center of the parabola to either side, the entire width of the water at the top will be \(2x\). The trough is \( 3 \) feet (\( 36 \) inches) long, so the surface area will then be:\[ \begin{array}{rcl} \text{Area} & = & l \cdot w \\[6pt] & = & 36 \cdot 2x \\[6pt] & = & 72x \\[6pt] & = & 72\sqrt{2y} \\[6pt] \end{array} \nonumber \]This example illustrates two important points:
- When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
- 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. Recall that such functions are called invertible functions , and we use the notation \( f^{-1}(x)\).
As a reminder: \(f^{-1}(x)\) is not the same as the reciprocal of the function \(f( x )\). This use of "–1" is reserved to denote inverse functions. To denote the reciprocal of a function \(f( x )\), we would need to write \( \left( f( x ) \right)^{-1} = \frac{1}{f( x )} \).
An important relationship between inverse functions is that they "undo" each other. If \(f^{-1}\) is the inverse of a function \(f\), then \(f\) is the inverse of the function \(f^{-1} \). In other words, whatever the function \(f\) does to \(x\), \(f^{-1}\) undoes it - and vice-versa. More formally, we write\[f^{-1}\left(f( x )\right) = x, \text{ for all }x\text{ in the domain of }f \nonumber \]and\[ f\left(f^{-1}( x )\right) = x, \text{ for all }x\text{ in the domain of } f^{-1}. \nonumber \]
Two functions, \(f\) and \(g\), are inverses of one another if for all \(x\) in the domain of \(f\) and \(g\).\[g( f( x ) )=f( g( x ) )=x. \nonumber \]
Show that \(f( x ) = \frac{1}{x+1} \) and \(f^{-1}( x )= \frac{1}{x} -1\) are inverses, for \(x \neq 0,-1\).
- Solution
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We must show that \(f^{-1} ( f( x ) )=x\) and \(f( f^{-1} ( x ) )=x\).\[ \begin{array}{rcl} f^{-1} (f(x)) & = & f^{-1} \left( \dfrac{1}{x+1} \right) \\[6pt] & = & \dfrac{1}{\frac{1}{x+1}} - 1 \\[6pt] & = & (x+1)-1 \\[6pt] & = & x \\[6pt] & & \\[6pt] f( f^{-1} (x)) & = & f\left( \dfrac{1}{x} - 1 \right) \\[6pt] & = & = \dfrac{1}{\left( \frac{1}{x} - 1 \right) +1} \\[6pt] & = & \dfrac{1}{\frac{1}{x}} \\[6pt] & = & x \\[6pt] \end{array} \nonumber \]Therefore, \(f( x ) = \frac{1}{x+1}\) and \(f^{-1} ( x ) = \frac{1}{x} -1\) are inverses.
Show that \(f( x )= \frac{x+5}{3} \) and \(f^{-1}( x )=3x-5\) are inverses.
Find the inverse of the function \(f(x)=5 x^3 +1\).
- Solution
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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 \(x\).\[ \begin{array}{rrcl} & y & = & 5 x^3 +1 \\[6pt] \implies & y - 1 & = & 5 x^3 \\[6pt] \implies & \dfrac{y - 1}{5} & = & x^3 \\[6pt] \implies & \sqrt[3]{\dfrac{y - 1}{5}} & = & x \\[6pt] \implies & f^{-1}(x) & = & \sqrt[3]{\dfrac{x-1}{5}} \\[6pt] \end{array} \nonumber \]
Look at the graph of \(f\) and \(f^{-1} \) in Figure \( \PageIndex{ 4 } \). Notice that the two graphs are symmetric about the line \(y=x\). This is always the case when graphing a function and its inverse function.
Also, since the method involved interchanging \(x\) and \(y\), notice corresponding points. If \((a,b)\) is on the graph of \(f\), then \((b,a)\) is on the graph of \(f^{-1} \). Since \((0,1)\) is on the graph of \(f\), then \((1,0)\) is on the graph of \(f^{-1} \). Similarly, since \((1,6)\) is on the graph of \(f\), then \((6,1)\) is on the graph of \(f^{-1} \).
Figure \( \PageIndex{ 4 } \)
Find the inverse function of \(f(x)= \sqrt[3]{x+4} \).
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 an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.
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.
- Restrict the domain by determining a domain on which the original function is one-to-one.
- Replace \(f(x)\) with \(y\).
- Solve for \(x\) interchange \( x \) and \( y\).
- Rename the function \(f^{-1} (x)\).
- Revise the formula for \(f^{-1} (x)\) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.
Find the inverse function of \(f:\)
- \(f(x)= (x-4)^2, \, x \geq 4\)
- \(f(x)= (x-4)^2, \, x \leq 4\)
- Solution
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The original function \(f(x)= (x-4)^2\) is not one-to-one, but the function is restricted to a domain of \(x \geq 4\) or \(x \leq 4\) on which it is one-to-one. See Figure \( \PageIndex{ 5 } \).
Figure \( \PageIndex{ 5 } \)To find the inverse, start by replacing \(f(x)\) with the simple variable \(y\).\[ \begin{array}{rrclcl} & y & = & (x-4)^2 & \quad & \left( \text{letting }y = f(x) \right) \\[6pt] \implies & \pm \sqrt{ y } & = & x-4 & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \implies & 4 \pm \sqrt{ y } & = & x & \quad & \left( \text{adding }4 \text{ to both sides} \right) \\[6pt] \implies & 4 \pm \sqrt{x} & = & y & \quad & \left( \text{exchanging }y\text{ and }x \right) \\[6pt] \end{array} \nonumber \]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 \(x\) and \(y\) for the original \(f(x)\), we need to look at the domain of the original function. When we reversed the roles of \(x\) and \(y\), this gave us the values \(y\) could assume. For this function, \(x \geq 4\), so for the inverse, we should have \(y \geq 4\), which is what our inverse function gives.
- The domain of the original function was restricted to \(x \geq 4\), so the outputs of the inverse need to be the same, \(f( x ) \geq 4\), and we must use the "+" case:\[f^{-1}(x)=4+\sqrt{ x }. \nonumber \]
- The domain of the original function was restricted to \(x \leq 4\), so the outputs of the inverse need to be the same, \(f( x ) \leq 4\), and we must use the "–" case:\[f^{-1} (x)=4-\sqrt{ x }. \nonumber \]
On the graphs in Figure \( \PageIndex{ 6 } \), we see the original function from Example \( \PageIndex{ 3 } \) graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line \(y=x\). The coordinate pair \((4,0)\) is on the graph of \(f\) and the coordinate pair \((0,4)\) is on the graph of \(f^{-1} \). For any coordinate pair, if \(( a,b )\) is on the graph of \(f\), then \(( b,a )\) is on the graph of \(f^{-1} \). Finally, observe that the graph of \(f\) intersects the graph of \(f^{-1}\) on the line \(y=x\). Points of intersection for the graphs of \(f\) and \(f^{-1}\) will always lie on the line \(y=x\).
Figure \( \PageIndex{ 6 } \)
Restrict the domain and then find the inverse of\[f(x)= (x-2)^2 -3. \nonumber \]
- Solution
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We can see this is a parabola with vertex at \((2,–3)\) 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 \(x \geq 2\).
To find the inverse, we repeat what we did in Example \( \PageIndex{ 3 } \).\[ \begin{array}{rrclcl} & y & = & (x - 2)^2 - 3 & \quad & \left( \text{substituting }y = f(x) \right) \\[6pt] \implies & y + 3 & = & (x - 2)^2 & \quad & \left( \text{adding }3\text{ to both sides} \right) \\[6pt] \implies & \pm \sqrt{y + 3} & = & x - 2 & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \implies & 2 \pm \sqrt{y + 3} & = & x & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt] \implies & 2 \pm \sqrt{x + 3} & = & y & \quad & \left( \text{interchanging }x\text{ and }y \right) \\[6pt] \end{array} \nonumber \]Now we need to determine which case to use. Because we restricted our original function to a domain of \(x \geq 2\), the outputs of the inverse should be the same, telling us to utilize the "+" case. Therefore,\[f^{-1}(x)=2+\sqrt{ x+3 }. \nonumber \]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.
Notice that we arbitrarily decided to restrict the domain on \(x \geq 2\). We could just have easily opted to restrict the domain on \(x \leq 2\), in which case \(f^{-1}(x)=2-\sqrt{ x+3 } \). Observe the original function graphed on the same set of axes as its inverse function in Figure \( \PageIndex{ 7 } \). Notice that both graphs show symmetry about the line \(y=x\). The coordinate pair \(( 2,-3 )\) is on the graph of \(f\) and the coordinate pair \(( -3,2 )\) is on the graph of \(f^{-1} \). Observe from the graph of both functions on the same set of axes that\[\text{domain of }f=\text{range of } f^{-1} =[ 2, \infty ) \nonumber \]and\[\text{domain of } f^{-1} =\text{range of }f=[ –3, \infty ). \nonumber \]Finally, observe that the graph of \(f\) intersects the graph of \(f^{-1}\) along the line \(y=x\).
Figure \( \PageIndex{ 7 } \)
Find the inverse of the function \(f(x)= x^2 +1\), on the domain \(x \geq 0\).
Solving Applications Involving the Inverse of Polynomial 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.
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\[V= \dfrac{2}{3} \pi r^3. \nonumber \]Find the inverse of this function. Then use the inverse function to calculate the radius of such a mound of gravel measuring \( 100 \) cubic feet. Use \(\pi \approx 3.14\).
- Solution
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Start with the given function for \(V\). Notice that the meaningful domain for the function is \(r \geq 0\) 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 \(V \geq 0\). Solve for \(r\) in terms of \(V\), using the method outlined previously.\[ \begin{array}{rrcl} & V & = & \dfrac{2}{3} \pi r^3 \\[6pt] \implies & \dfrac{3}{2} V & = & \pi r^3 \\[6pt] \implies & \dfrac{3}{2 \pi} V & = & r^3 \\[6pt] \implies & \sqrt[3]{\dfrac{3}{2 \pi} V} & = & r \\[6pt] \end{array} \nonumber \]This is the result stated in the section opener. Now evaluate this for \(V=100\) and \(\pi \approx 3.14\).\[ \begin{array}{rcl} r & = & \sqrt[3]{\dfrac{3V}{2\pi}} \\[6pt] & \approx & \sqrt[3]{\dfrac{3(100)}{2 (3.14)}} \\[6pt] & \approx & \sqrt[3]{47.7707} \\[6pt] & \approx & 3.63 \\[6pt] \end{array} \nonumber \]Therefore, the radius is about \( 3.63 \) ft.
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.
The function \(C = \frac{20+0.4n}{100+n}\) represents the concentration \(C\) of an acid solution after \(n\) 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 \(n\) in terms of \(C\). Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.
- Solution
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We first want the inverse of the function.\[ \begin{array}{rrcl} & C & = & \dfrac{20 + 0.4n}{100 + n} \\[6pt] \implies & C(100 + n) & = & 20 + 0.4n \\[6pt] \implies & 100C + Cn & = & 20 + 0.4n \\[6pt] \implies & 100C - 20 & = & 0.4n - Cn \\[6pt] \implies & 100C - 20 & = & (0.4 - C)n \\[6pt] \implies & \dfrac{100C - 20}{0.4 - C} & = & n \\[6pt] \end{array} \nonumber \]Now evaluate this function for \(C= 0.35\) (35%).\[ n = \dfrac{100(0.35)-20}{0.4-0.35} = \dfrac{15}{0.05} = 300. \nonumber \]We can conclude that 300 mL of the 40% solution should be added.
Find the inverse of the function \(f(x)= \dfrac{x+3}{x-2} \).
Access these online resources for additional instruction and practice with inverses and radical functions.