In this section we will extend our previous work with functions to officially include radicals. A lot of the material in this section will feel like a review of material from early in the course; however, after a few brief definitions, theorems, and examples at this "review" level, we will increase the level to be a bit more challenging.
Recall from Algebra that if a function is defined by a radical expression, we call it a radical function. In particular,
the square root function is \(f(x)=\sqrt{x}\).
the cube root function is \(f(x)=\sqrt[3]{x}\).
For completeness, we include the official definition of a radical function here.
Definition: Radical Function
A radical function is a function that is defined by a radical expression.
Let's take a moment to review how to evaluate radical functions.
We follow the same process as in part a.\[ \begin{array}{rclcl} f(-2) & = & \sqrt{2 \cdot (-2)-1} & \quad & \left( \text{substituting} \right) \\[6pt] & = & \sqrt{-4-1} & \quad & \left( \text{simplifying} \right) \\[6pt] & = & \sqrt{-5} & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]Since the square root of a negative number is not a real number, the function does not have a (real) value at \(x=-2\).
Example \( \PageIndex{ 1b } \) illustrates an important point - when working with functions (radical or not), we are assuming that the number system we're interested in is the real number system. Therefore, we do not want imaginary numbers as function values.
Checkpoint \(\PageIndex{1}\)
For the function \(f(x)=\sqrt{3 x-2}\), find
\(f(6)\)
\(f(0)\)
Answers
\(f(6)=4\)
no value at \(x=0\)
Given a cube root function, we follow the same procedure.
As we did with part a, we substitute.\[ \begin{array}{rclcl} g(-2) & = & \sqrt[3]{-2 - 6} & \quad & \left( \text{substituting} \right) \\[6pt] & = & \sqrt[3]{-8} & \quad & \left( \text{simplifying} \right) \\[6pt] & = & -2 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]
As Example \( \PageIndex{ 2b } \) shows, one of the major differences between a square root function and a cube root function is that we can evaluate the cube root of a negative number. In fact, as you (hopefully) remember from Algebra, the domain of a radical function with an odd index is all real numbers, while the domain of an even-indexed radical function is restricted so that the radicand (the argument of the radical function) is non-negative. We will write this as a theorem in a little bit.
Checkpoint \(\PageIndex{2}\)
For the function \(g(x)=\sqrt[3]{3 x-4}\), find
\(g(4)\)
\(g(1)\)
Answers
\(g(4)=2\)
\(g(1)=-1\)
Again, as you should recall from Algebra, radical functions need not be restricted to square roots and cube roots.
\[ \begin{array}{rclcl} f(4) & = & \sqrt[4]{5 \cdot (-12) - 4} & \quad & \left( \text{substituting} \right) \\[6pt] & = & \sqrt[4]{-60 - 4} & \quad & \left( \text{simplifying} \right) \\[6pt] & = & \sqrt[4]{-64} & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]Since the fourth root of a negative number is not a real number, the function does not have a (real) value at \(x=-12\).
Checkpoint \(\PageIndex{3}\)
For the function \(f(x)=\sqrt[4]{3 x+4}\), find
\(f(4)\)
\(f(-1)\)
Answers
\(f(4)=2\)
\(f(-1)=1\)
The Domain of a Radical Function
To find the domain of a radical function, we use our properties of radicals from Algebra. For a radical with an even index, we said the radicand had to be greater than or equal to zero. This is because even roots of negative numbers are not real numbers. For an odd index, the radicand can be any real number. We restate the properties here for reference.
Theorem: Properties of \(\sqrt[n]{a}\)
When \(n\) is an even number and:
\(a \geq 0\), then \(\sqrt[n]{a}\) is a real number.
\(a < 0\), then \(\sqrt[n]{a}\) is not a real number.
When \(n\) is an odd number, \(\sqrt[n]{a}\) is a real number for all values of \(a\).
Therefore, to find the domain of a radical function with an even index, we set the radicand to be greater than or equal to zero. For an odd index radical, the radicand can be any real number. Again, for completeness, we include this statement as a theorem.
Theorem: Domain of a Radical Function
When the index of the radical is even, the domain of the radical function consists of all real numbers for which the radicand is greater than or equal to zero. When the index of the radical is odd, the domain of a radical function is all real numbers.
Example \(\PageIndex{4}\)
Find the domain of the function, \(f(x)=\sqrt{3 x-4}\). Write the domain in interval notation.
Solution
Since the function, \(f(x)=\sqrt{3 x-4}\) has a radical with an index of \(2\), which is even, we know the radicand must be greater than or equal to \(0\).\[ \begin{array}{rrclcl} & 3x - 4 & \geq & 0 & \quad & \left( \text{setting the radicand greater than or equal to }0 \right) \\[6pt] \implies & 3x & \geq & 4 & \quad & \left( \text{adding }4\text{ to both sides} \right) \\[6pt] \implies & x & \geq & \dfrac{4}{3} & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt] \end{array} \nonumber \]The domain of \(f(x)=\sqrt{3 x-4}\) is all values \(x \geq \frac{4}{3}\), and we write it in interval notation as \(\left[\frac{4}{3}, \infty\right)\).
Checkpoint \(\PageIndex{4}\)
Find the domain of the function, \(f(x)=\sqrt{6 x-5}\). Write the domain in interval notation.
Answer
\(\left[\frac{5}{6}, \infty\right)\)
Up to this point in this section, we really have not stated anything new. Everything we have covered should feel like a repeat from your previous Algebra classes. Now we begin to improve upon these older skills to introduce examples more relevant to Calculus.
Example \(\PageIndex{5}\)
Find the domain of the function, \(g(x)=\sqrt{\frac{6}{x-1}}\). Write the domain in interval notation.
Solution
Since the function, \(g(x)=\sqrt{\frac{6}{x-1}}\), has a radical with an even index, we know the radicand must be greater than or equal to \(0\). Moreover, since the numerator is a nonzero constant, the radicand cannot be zero.
For \(\frac{6}{x-1}\) to be greater than zero, the denominator must be positive (since the numerator is positive). Therefore, we set \(x-1 > 0\) and solve. This yields \( x > 1 \). So far, our candidate domain is \( \left( 1,\infty \right) \).
An additional consideration when dealing with functions involving fractions is that the denominator cannot be zero. Therefore, we know \( x \neq 1 \); however, since \( 1 \) is not in the interval \( \left( 1,\infty \right) \), we don't have to worry about this extra restriction.
Putting this together we get the domain is \((1, \infty)\).
Checkpoint \(\PageIndex{5}\)
Find the domain of the function, \(f(x)=\sqrt{\frac{4}{x+3}}\). Write the domain in interval notation.
Answer
\((-3, \infty)\)
Example \(\PageIndex{6}\)
Find the domain of the function, \(h(x)=\sqrt[6]{-21x^2 + 2x + 8}\). Write the domain in interval notation.
Solution
The even index on this radical function tells us that the radicand must be greater than or equal to zero. Therefore, we need to solve\[ -21x^2 +2x +8 \geq 0. \nonumber \]Proceeding as we did when solving quadratic inequalities, we factor this quadratic to get\[ \left( -3x + 2 \right)\left( 7x + 4 \right) \geq 0. \nonumber \]We then sketch a graph of the quadratic expression on the left side of the inequality to find when the curve is at or above the \( x \)-axis.
We can see from the graph that the inequality is satsified when \( -\frac{4}{7} \leq x \leq \frac{2}{3} \). Hence, the domain of the radical function is \( \left[ -\frac{4}{7}, \frac{2}{3} \right] \).
Checkpoint \(\PageIndex{6}\)
Find the domain of the function, \(f(x)=\sqrt{x^3 + x^2 - 4x - 4}\). Write the domain in interval notation.
Answer
\([-2, -1] \cup [2, \infty)\)
Example \( \PageIndex{ 6 } \) and Checkpoint \( \PageIndex{ 6 } \) illustrate the need to keep our skills with the material we previously covered in this course sharp. This is the way of Calculus - once you have learned a skill, it is likely going to come into play at some point in the future.
Example \( \PageIndex{ 7 } \): Finding the Domain of a Radical Function Composed with a Rational Function
Find the domain of the function \(f(x)=\sqrt{ \frac{ (x+2)(x-3) }{ x-1 } } \).
Solution
Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where \( \frac{(x+2)(x-3)}{x-1} \geq 0\). 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, we will continue to reinforce the graphing skills we learned previously and choose to sketch the rational function (see Figure \( \PageIndex{ 1 } \)).
Figure \( \PageIndex{ 1 } \)
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 \(f( x )\) will be defined. \(f( x )\) has domain \( -2 \leq x < 1\) or \( x \geq 3\), or in interval notation, \([-2,1) \cup [3, \infty )\).
Domains of Non-Radical Functions Containing Radicals
In Calculus, you often work with functions that contain radical expressions in addition to containing non-radical expressions. Technically speaking, these are not called radical functions - they are simply referred to as compositions of functions (a catchall phrase encompassing most functions). Examples of these styles of functions are:
\(f(x) = 3x \sqrt[3]{2-x}\)
\(g(x) = \sqrt{2-\sqrt[4]{x+3}}\)
\(k(x) = \frac{2x}{\sqrt{x^2 - 1}}\)
Being able to identify the domains of functions like these is going to be a necessary skill for success in Calculus.
To find the domains of functions like these, we use a method akin to the Order of Operations - working from the innermost function to the outermost. It is often easier to use the method of sign charts when finding the domains of composite functions.
When creating a sign chart for a given function, we are looking for when a denominator becomes zero, when an even-indexed radical has a negative as its radicand, or when the argument of a logarithm (which we haven't encountered yet) is zero or negative. There are other, more rare, cases where domain issues occur, but we will approach those only as needed.
Let's spend some time investigating how such domains can be determined.
Example \( \PageIndex{8} \)
State the domains of the following functions.
\(f(x) = 3x \sqrt[3]{2-x}\)
\(g(x) = \sqrt{2-\sqrt[4]{x+3}}\)
\(k(x) = \frac{2x}{\sqrt{x^2 - 1}}\)
Solutions
As far as domain is concerned, \(f(x)\) has no denominators and no even roots, which means there are no domain issues. Hence, the domain of this function is \( \left( -\infty, \infty \right) \).
In \(g(x) = \sqrt{2-\sqrt[4]{x+3}}\), we have two radicals both of which are even-indexed. To satisfy the innermost radical, \(\sqrt[4]{x+3}\), we require \(x+3 \geq 0\) or \(x \geq -3\). Thus, our candidate domain of \( g(x) \) is \( \left[ -3,\infty \right) \). This is the largest possible domain for the function \( g \); however, we have more to consider, and this domain might shrink.
"Zooming out" in our function, we see we need \(2-\sqrt[4]{x+3} \geq 0\). While it may be tempting to write this as \(2 \geq \sqrt[4]{x+3}\) and take both sides to the fourth power, there are times when this technique will produce erroneous results. Instead, we solve \(2-\sqrt[4]{x+3} \geq 0\) using a sign chart. We take a moment to find out when \(2-\sqrt[4]{x+3} = 0\).\[ \begin{array}{rrclcl} & 2-\sqrt[4]{x+3} & = & 0 & \quad & \\[6pt] \implies & 2 & = & \sqrt[4]{x+3} & \quad & \left( \text{adding }\sqrt[4]{x+3}\text{ to both sides} \right) \\[6pt] \implies & 2^4 & = & \left(\sqrt[4]{x+3}\right)^4 & \quad & \left( \text{raising both sides to the fourth power} \right) \\[6pt] \implies & 16 & = & x + 3 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & 13 & = & x & \quad & \left( \text{subtracting }3\text{ from both sides} \right) \\[6pt] \end{array} \nonumber \]One thing to be careful of throughout all of Mathematics is raising both sides of an equation to an even power - this sometimes introduces extraneous solutions. Therefore, we need to check to see if \(x=13\) is an extraneous solution. Substituting \(x=13\) into the expression \(\sqrt{2-\sqrt[4]{x+3}}\) does check since\[ 2-\sqrt[4]{x+3} = 2 - \sqrt[4]{13+3} = 2 - \sqrt[4]{16} = 2 - 2 = 0. \nonumber \]Remember, we are currently trying to figure out when \(2-\sqrt[4]{x+3} \geq 0\). We make a sign chart with marked with \( x = 13 \) (where the expression on the left side of the inequality is zero) and, since the candidate domain is \( \left[ -3,\infty \right) \), we start our number line at \( x = -3 \).
We find \(2-\sqrt[4]{x+3} \geq 0\) on \([-3,13]\) so this is the domain of \(g\).
To find the domain of \(k\), we have both an even root and a denominator to concern ourselves with. To satisfy the square root, \(x^2 - 1 \geq 0\). Setting \(r(x) = x^2-1\), we find the zeros of \(r\) to be \(x = \pm 1\), and we find the sign chart of \(r\) to be
We find \(x^2 - 1 \geq 0\) for \((-\infty, -1] \cup [1, \infty)\). This is our candidate domain, which might shrink as we look further into the problem.
To keep the denominator of \(k(x)\) away from zero, we set \(\sqrt{x^2-1} = 0\). We leave it to the reader to verify the solutions are \(x = \pm 1\), both of which must be excluded from the domain. Hence, the domain of \(k\) is \((-\infty, -1) \cup (1,\infty)\).
Graphing Higher-Indexed Radical Functions
We already learned the graphs of the square root and cube root functions, however, we have yet to graph higher-indexed radical functions.
It is worth remarking that we could define \(f(x) = \sqrt[n]{x}\) functionally as the inverse of \(g(x) = x^n\) with the stipulation that when \(n\) is even, the domain of \(g\) is restricted to \([0, \infty)\). From what we know about \(g(x) = x^n\) from our investigations of polynomials and their inverses, we can produce the graphs of \(f(x) = \sqrt[n]{x}\) by reflecting the graphs of \(g(x) = x^n\) across the line \(y=x\). Below are the graphs of \(y=\sqrt{x}\), \(y=\sqrt[4]{x}\) and \(y=\sqrt[6]{x}\). The point \((0,0)\) is indicated as a reference. The axes are hidden so we can see the vertical steepening near \(x=0\) and the horizontal flattening as \(x \to \infty\).
The odd-indexed radical functions also follow a predictable trend - steepening near \(x = 0\) and flattening as \(x \to \pm \infty\). In the exercises, you’ll have a chance to graph some basic radical functions using transformations.
We know the base graph of this function (it is in the images above), so we just need to apply the transformations. We begin by modifying the function to get it into an appropriate form.\[ f(x) = -\dfrac{1}{2}\sqrt[6]{2x + 8} + 1 = -\dfrac{1}{2} \sqrt[6]{2\left( x + 4 \right)} + 1.\nonumber \]Reading from left to right, the base graph of \( y = \sqrt[6]{x} \):
gets reflected about the \( x \)-axis (\( y = \textcolor{red}{-}\sqrt[6]{x} \))
gets compressed vertically by a factor of \( \frac{1}{2} \) (\( y = -\textcolor{red}{\frac{1}{2}}\sqrt[6]{x} \))
gets compressed horizontally by a factor of \( \frac{1}{2} \) (\( y = -\frac{1}{2}\sqrt[6]{\textcolor{red}{2}x} \))
gets shifted \(4\) units to the left (\( y = -\frac{1}{2}\sqrt[6]{2\left(x + \textcolor{red}{4}\right)} \))
gets shifted \(1\) unit up (\( y = -\frac{1}{2}\sqrt[6]{2\left(x + 4\right)} + \textcolor{red}{1} \))
The graph of \( f(x) \) is shown below.
Applications Involving Radicals
Radical functions often arise in Calculus as a result of a process called optimization. The following example illustrates one such optimization problem (but we are allowing ourselves technology because we don't have Calculus yet).
Example \( \PageIndex{10} \)
Carl wishes to get high speed internet service installed in his remote Sasquatch observation post located \(30\) miles from Route \(117\). The nearest junction box is located \(50\) miles down the road from the post, as indicated in the diagram below. Suppose it costs \(\$ 15\) per mile to run cable along the road and \(\$ 20\) per mile to run cable off of the road.
Express the total cost \(C\) of connecting the junction box to the post as a function of \(x\), the number of miles the cable is run along Route \(117\) before heading off road directly towards the post. Determine a reasonable applied domain for the problem.
Use graphing technology to graph \(y=C(x)\) on its domain. What is the minimum cost? How far along Route \(117\) should the cable be run before turning off the road?
Solution
The cost is broken into two parts: the cost to run cable along Route \(117\) at \(\$15\) per mile, and the cost to run it off road at \(\$20\) per mile. Since \(x\) represents the miles of cable run along Route \(117\), the cost for that portion is \(15x\). From the diagram, we see that the number of miles the cable is run off road is \(z\), so the cost of that portion is \(20z\). Hence, the total cost is \(C = 15x + 20z\).
Our next goal is to determine \(z\) as a function of \(x\). The diagram suggests we can use the Pythagorean Theorem to get\[y^2 + 30^2 = z^2.\nonumber \]But we also see \(x+y = 50\), so that \(y=50-x\). Hence,\[z^2 = (50-x)^2+900.\nonumber \]Solving for \(z\), we obtain\[z = \pm \sqrt{(50-x)^2+900}.\nonumber \]Since \(z\) represents a distance, we choose \(z = \sqrt{(50-x)^2+900}\) so that our cost as a function of \(x\) only is given by\[C(x) = 15x + 20\sqrt{(50-x)^2+900}.\nonumber\]From the context of the problem, we have the applied domain of \( x \in \left[ 0,50 \right]\).
Graphing \(y=C(x)\) using Desmos, we find the relative minimum (which is also the absolute minimum in this case) to two decimal places to be \((15.98, 1146.86)\).
Here the \(x\)-coordinate tells us that in order to minimize cost, we should run \(15.98\) miles of cable along Route \( 117 \) and then turn off of the road and head towards the outpost. The \(y\)-coordinate tells us that the minimum cost, in dollars, to do so is \(\$1146.86\). The ability to stream live SasquatchCasts? Priceless.
