ORCCA Introduction to Radical Functions

Section 13.1 Introduction to Radical Functions

We learned the basics of square roots in Section 9.1. The study of radicals is much broader than our first attempt covered and we need to expand our investigation. To do so, we will first look at an example that makes use of a topic covered in Section 9.2.

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A #10 washer has a 5.6 mm inner diameter and is 1.2 mm thick. We will let

$d$ represent the outer diameter, measured in mm.

Figure 13.1.1. A Diagram of a #10 Washer

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The amount of steel,

$M\text{,}$ in mg that it takes to make the washer with outer diameter

$d$ is approximated by the formula

$\begin{equation*} M=7.59d^2-238 \end{equation*}$

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Note that if you know the value of

$M$ ahead of time, this is a quadratic equation. We will now solve the equation for

$d$ using the square root method.

$\begin{align*} M\amp=7.59d^2-238\\ M\addright{238}\amp=7.59d^2\\ \divideunder{M+238}{7.59}\amp=d^2 \end{align*}$

$\begin{align*} d\amp=\sqrt{\frac{M+238}{7.59}}\amp\amp\text{or}\amp d\amp=-\sqrt{\frac{M+238}{7.59}} \end{align*}$

Since we know that the diameter cannot really be negative, our formula for $d$ must be

$\begin{equation*} d=\sqrt{\frac{M+238}{7.59}} \end{equation*}$

This formula finds the diameter that the washer must be when you input the amount of steel used, $M\text{.}$ Figure 13.1.2 shows a graph of this relationship.

Figure 13.1.2. A Graph of

$d=\sqrt{\frac{M+238}{7.59}}$

We also know that $M$ cannot be negative, so we will cut the graph to begin at $M=0\text{.}$ This formula tells us that if we plan on using, for example, 1000 mg of steel (about as much as in a large paper clip) that we can find the the outer diameter for the washer that will be created:

$\begin{align*} d\amp=\sqrt{\frac{\substitute{1000}+238}{7.59}}\\ \amp\approx12.8 \end{align*}$

So the washer's outer diameter must be about 12.8 mm for it to have a mass of 1000 mg.

Figure 13.1.3. A Revised Graph of

$d=\sqrt{\frac{M+238}{7.59}}$

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Note that the vertical intercept of the graph is

$(0,5.6)\text{,}$ which says that a washer that uses no steel at all (is that really a washer?) would have an outer diameter of 5.6 mm. This is the “smallest” possible washer with an inner diameter of 5.6 mm, even though it would technically be massless. Perhaps the implied domain of this function should be

$(0,\infty)$ to exclude

$0$ mass.

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Square roots often appear when we consider formulas from geometry like the washer problem, and they also show up in topics like in statistics (where

$\sigma=\sqrt{\frac{1}{n}\sum\limits_{i=1}^n (x_i-\mu)^2}$ finds the standard deviation), in chemistry (where

$v_{rms}=\sqrt{\frac{3RT}{M_m}}$ finds the velocity of a particle), and in physics (where

$m(v)=\frac{m_0}{\sqrt{1-\sfrac{v^2}{c^2}}}$ finds the mass of an object as its velocity nears the speed of light). There are many more examples to give, but we need a firmer understanding of radicals to properly study these things, so it's time to venture into deeper waters.

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Subsection 13.1.1 The Square Root Function

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Example 13.1.4.

Gilberto is an artist who etches designs into square copper plates of different sizes. Customers can order the size they would like.

Build a function that calculates the length of a plate's side given its area. Explore the function with a table and graph.
One customer ordered a plate with an area of $6.25$ square feet. Calculate the length of its side.
Find the domain and range of the function from Part a.

Explanation

We know the formula to calculate a square's area is $A=l^2\text{,}$ where $A$ stands for a square's area, and $l$ is the length of the square's side. To build a function to calculate $l\text{,}$ we solve for $l$ in the formula:

\begin{align*} A\amp=l^2 \end{align*}

\begin{align*} l\amp=\sqrt{A}\amp\amp\text{or} \amp l\amp=-\sqrt{A} \end{align*}

The formula is $l=\sqrt{A}\text{.}$ We don't consider the negative solution in this context since negative length doesn't make sense. Since $l$ depends on $A\text{,}$ we can use function notation and write $f(A)=\sqrt{A}\text{.}$ We will make a table, plot points, and look at the graph of $l=\sqrt{A}\text{.}$

Table 13.1.5. Values of $f(A)=\sqrt{A}$

$A$	$f(A)$	Points on the Curve
$0$	$\sqrt{0}=0$	$(0,0)$
$1$	$\sqrt{1}=1$	$(1,1)$
$4$	$\sqrt{4}=2$	$(4,2)$
$6.25$	$\sqrt{6.25}=2.5$	$(6.25,2.5)$
$9$	$\sqrt{9}=3$	$(9,3)$

Figure 13.1.6. Graph of $l=\sqrt{A}$

The point $(6.25,2.5)$ implies that a square plate with an area of $6.25$ square feet would have a length of $2.5$ feet on each side.
According to the graph, the function's domain is $[0,\infty)$ because the graph goes forever to the right from $A=0\text{.}$ This should make some sense because you cannot take the square root of a negative number. The function's range is $[0,\infty)$ because the graph seems to go up forever starting at $l=0\text{.}$ This should also make sense because a square root never gives you a negative number as an answer.

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Fact 13.1.7. Domain of a Square Root Functions.

To algebraically find the domain of a square root function, set the radicand (the expression under the radical) greater than or equal to $0$ and solve for the variable. The solution set to that inequality is the domain of the function.

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Example 13.1.8.

Algebraically find the domain of the function $g$ where $g(x)=\sqrt{2x-4}+1$ and then find the range by making a graph.

Explanation

Using Fact 13.1.7 to find the function's domain, we set the radicand greater than or equal to zero and solve:

\begin{align*} 2x-4\amp\ge0\\ 2x\amp\ge4\\ x\amp\ge2 \end{align*}

The function's domain is $[2,\infty)$ in interval notation.

To find the function's range, we use point plotting to graph the function and then look at its graph. The graph shows that the function's range is $[1,\infty)\text{.}$ The graph also verifies the function's domain is indeed $[2,\infty)\text{.}$

Figure 13.1.9. Graph of $g(x)=\sqrt{2x-4}+1$

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Example 13.1.10.

Algebraically find the domain and graphically find the range of the function $h$ where $h(x)=2-3\sqrt{8-5x}\text{.}$

Explanation

To find the function's domain, we set the radicand to be greater than or equal to zero:

\begin{align*} 8-5x\amp\ge0\\ -5x\amp\ge-8\\ \divideunder{-5x}{-5}\amp\mathbin{\highlight{\le}}\divideunder{-8}{-5}\\ x\amp\le\frac{8}{5} \end{align*}

So, the function's domain is $\left(-\infty,\frac{8}{5}\right]\text{.}$ The $2$ and $3$ in the function do not play a role in the domain, although they do alter the range, which we will find now by making a graph.

From the graph, we can see that the range is all numbers below (or equal to) the $y$-value $2\text{.}$ In interval notation, this would be written $(-\infty,2]\text{.}$

Figure 13.1.11. Graph of $h(x)=2-3\sqrt{8-5x}$

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Example 13.1.12.

When an object is dropped, the time it takes to hit the ground, $t\text{,}$ in seconds, can be modeled by

$\begin{equation*} t=\sqrt{\frac{d}{16}} \end{equation*}$

where $d$ stands for the initial height of the object in feet. Use algebra to answer the following questions.

In a science experiment, Amaka's class drops a beanbag from the top of a $100$ -foot-tall building. How long will it take for the beanbag to hit the ground?
Her class then goes to a second building, drops the beanbag from the top, and uses a stopwatch to measures the time it takes to hit the ground. If it takes $3$ seconds for the beanbag to hit the ground, how tall is the building?

Explanation

To find how long it will take for the object to hit the ground, we need to plug in $100$ for the distance that it will fall,$d$ and then solve for $t\text{.}$

\begin{align*} t=\sqrt{\frac{d}{16}}\\ t\amp=\sqrt{\frac{100}{16}}\\ t\amp=\frac{\sqrt{100}}{sqrt{16}}\\ t\amp=\frac{10}{4}\\ t\amp=2.5 \end{align*}

This means it will take the beanbag $2.5$ seconds to hit the ground if it's released from the top of a $100$-foot-tall building.
This time we substitute $3$ for $t$ and solve for $d\text{:}$

\begin{align*} t=\sqrt{\frac{d}{16}}\\ 3\amp=\sqrt{\frac{d}{16}}\\ 3^2\amp=\left(\sqrt{\frac{d}{16}}\right)^2\\ 9\amp=\frac{d}{16}\\ 16\cdot 9\amp=\frac{d}{16}\cdot 16\\ 144\amp=d \end{align*}

This means the beanbag will fall approximately $144$ feet in $3$ seconds, so the second building is approximately $144$ feet tall.

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Subsection 13.1.2 The Distance Formula

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A square root is used in calculating the distance between two points on a coordinate plane. We learned the Pythagorean Theorem in Section 9.2.2. In a coordinate plane, we can use the Pythagorean Theorem to calculate the distance between any two points.

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Example 13.1.13.

Calculate the distance between $(2,3)$ and $(5,7)\text{.}$

Explanation

First, we will sketch a graph of those two points.

Figure 13.1.14. Calculating the Distance Between $(2,3)$ and $(5,7)$

To calculate the distance between $(2,3)$ and $(5,7)\text{,}$ we sketch a right triangle as in the figure and then use the Pythagorean Theorem:

\begin{align*} d^2\amp=(5-2)^2+(7-3)^2\\ d^2\amp=3^2+4^2\\ d^2\amp=9+16\\ d^2\amp=25\\ d\amp=\sqrt{25}\\ d\amp=5 \end{align*}

In conclusion, the distance between $(2,3)$ and $(5,7)$ is $5\text{.}$ Note that in our calculations, we didn't need to show $d=\pm\sqrt{25}$ because distance must have a positive value.

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With the same method, we can derive a formula to calculate the distance between any two points.

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Example 13.1.15.

With a generic first and second point, we will use subscripts to identify the first pair $(x_1,y_1)$ and the second pair $(x_2,y_2)\text{.}$ Calculate the distance between the generic points $(x_1,y_1)$ and $(x_2,y_2)\text{.}$

Explanation

First, we will sketch a graph of those two points. We will put the image from last example side by side with the new image, so it's clear that we are using the same method.

Figure 13.1.16. Calculating the Distance Between $(2,3)$ and $(5,7)$

Figure 13.1.17. Calculating the Distance Between $(x_1,y_1)$ and $(x_2,y_2)$

To calculate the distance between $\highlight{(x_1,y_1)}$ and $\highlight{(x_2,y_2)}\text{,}$ we sketch a right triangle as in the figure and then use Pythagorean Theorem:

\begin{align*} d^2\amp=(x_2-x_1)^2+(y_2-y_1)^2\\ d\amp=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{align*}

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Fact 13.1.18. The Distance Formula.

The distance between two points $\highlight{(x_1,y_1)}$ and $\highlight{(x_2,y_2)}\text{,}$ is given by the formula:

$\begin{align*} d\amp=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{align*}$

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With this formula, we can calculate the distance between two points without sketching a graph.

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Example 13.1.19.

Find the distance between $(-2,4)$ and $(5,-20)\text{.}$

Explanation

To calculate the distance between $(-2,4)$ and $(5,-20)\text{,}$ we use the distance formula. It's good practice to mark each value with the corresponding variables in the formula. Again, $(x_1,y_1)$ stands for the first point's coordinates, and $(x_2,y_2)$ stands for the second point's coordinates:

\begin{equation*} (\stackrel{x_1}{-2},\stackrel{y_1}{4}), (\stackrel{x_2}{5},\stackrel{y_2}{-20}) \end{equation*}

We have:

\begin{align*} d\amp=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d\amp=\sqrt{\left(\substitute{5}-\substitute{(-2)}\right)^2+\left(\substitute{(-20)}-\substitute{4}\right)^2}\\ d\amp=\sqrt{\left(7\right)^2+\left(-24\right)^2}\\ d\amp=\sqrt{49+576}\\ d\amp=\sqrt{625}\\ d\amp=25 \end{align*}

The distance between $(-2,4)$ and $(5,-20)$ is $25$ units.

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Warning 13.1.20.

Note that it's good practice to add parentheses around negative values when we do substitutions. For example, when we substitute $x$ with $-7$ in $x^2\text{,}$ we should write

$\begin{equation*} x^2=\substitute{(-7)}^2=49\text{ correct }\checkmark \end{equation*}$

We should not write

$\begin{equation*} x^2=\substitute{-7}^2=-49\text{ incorrect } \end{equation*}$

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Subsection 13.1.3 Cube Root Function

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The square of

$2$ is

$4\text{,}$ so the square root of

$4$ is

$2\text{.}$

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Similarly, the cube of

$2$ is

$8\text{,}$ so the cube root of

$8$ is

$2\text{.}$ We write

$\begin{equation*} \sqrt[3]{8}=2 \end{equation*}$

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It's helpful to memorize the first few perfect cube numbers and their cube roots:

$\begin{align*} \amp0^3=0\amp\sqrt[3]{0}=0\\ \amp1^3=1\amp\sqrt[3]{1}=1\\ \amp2^3=8\amp\sqrt[3]{8}=2\\ \amp3^3=27\amp\sqrt[3]{27}=3\\ \amp4^3=64\amp\sqrt[3]{64}=4\\ \amp5^3=125\amp\sqrt[3]{125}=5 \end{align*}$

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One major difference between a cube root and a square root is that we can find the cube root of negative numbers. For example:

$\begin{equation*} (-4)^3=-64\text{, so }\sqrt[3]{-64}=-4 \end{equation*}$

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However,

$\sqrt{-64}$ is non-real and in general we cannot take the square root of a negative number.

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Remark 13.1.21.

Many calculators don't have a cube root button. If yours does, it might look like $\sqrt[n]{\phantom{x}}$ and you will tell the calculator both to enter a number for the “ $n$ ” as well as the radicand. Many calculators also allow you to type something like root(3,8) for $\sqrt[3]{8}\text{,}$ for example.

Another way to calculate the cube root on a calculator is to use the exponent button (which is usually marked with the caret symbol, ^) with a reciprocal power. For example, to calculate $\sqrt[3]{8}\text{,}$ you may type 8^(1/3). We will explain why $\sqrt[3]{8}=8^{\frac{1}{3}}$ in Section 13.2. For now, just learn how to use a calculator to calculate the cube root of a given number.

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We can also estimate the value of a cube root, like

$\sqrt[3]{10}\text{,}$ by knowing the perfect cubes nearby:

$\begin{align*} \sqrt[3]{8}\amp=2\amp\sqrt[3]{10}\amp=\mathord{?}\amp\sqrt[3]{27}\amp=3 \end{align*}$

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Since

$10$ is between the perfect cubes

$8$ and

$27\text{,}$

$\sqrt[3]{10}$ must be between

$2$ and

$3\text{,}$ and closer to

$2\text{.}$ We can use a calculator to verify

$\sqrt[3]{10}\approx2.154$

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Let's build a table and graph the cube root function.

Table 13.1.22. Values of

$g(x)=\sqrt[3]{x}$

$x$	$g(x)=\sqrt[3]{x}$	Points on the Curve
$-8$	$\sqrt[3]{-8}=-2$	$(-8,-2)$
$-1$	$\sqrt[3]{-1}=-1$	$(-1,-1)$
$0$	$\sqrt[3]{0}=0$	$(0,0)$
$1$	$\sqrt[3]{1}=1$	$(1,1)$
$8$	$\sqrt[3]{8}=2$	$(8,2)$

Figure 13.1.23. Graph of

$g(x)=\sqrt[3]{x}$

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Both the domain and range of the cube root function are

$(-\infty,\infty)\text{.}$ Compare this with the domain and range of the square root function, which are each

$[0,\infty)\text{.}$ The reason for the difference is that we cannot take the square root of negative numbers, but we can take the cube root of negative numbers (and when we do, we get negative numbers as the output).

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Remark 13.1.24.

It is helpful to be able to quickly sketch the graphs of the following types of basic functions:

$\begin{align*} f(x)\amp=c\amp f(x)\amp=mx+b\amp f(x)\amp=a(x-h)^2+k\amp f(x)\amp=\abs{x}\\ f(x)\amp=\frac{1}{x}\amp f(x)\amp=\frac{1}{x^2}\amp f(x)\amp=\sqrt{x}\amp f(x)\amp=\sqrt[3]{x} \end{align*}$

Now with the graph of the cube root function, you have seen all of these shapes in this book.

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Example 13.1.25.

Nasim makes solid copper spheres for their grounding and healing properties. A sphere's radius can be calculated by the formula $r(V)=\sqrt[3]{\frac{3V}{4\pi}}\text{,}$ where $r(V)$ stands for the sphere's radius for a given volume $V\text{.}$ If Nasim uses $2$ cubic inches of copper per sphere, what diameter should he list on his website? Round your answer to two decimal places.

Explanation

First, to find the radius we will substitute $\substitute{2}$ in for $V\text{,}$ and we have:

\begin{align*} r(V)\amp=\sqrt[3]{\frac{3V}{4\pi}}\\ r(\substitute{2})\amp=\sqrt[3]{\frac{3(\substitute{2})}{4\pi}}\\ \amp\approx0.78 \end{align*}

The spheres will have a radius of approximately 0.78 in.

When we calculate $\sqrt[3]{\frac{3(2)}{4\pi}}$ with a calculator, we enter (3*2/(4π))^(1/3). To find the diameter, we multiply the radius by $2$ to get 1.56 in. Nasim can advertise the spheres to be 1.56 inches in diameter.

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Subsection 13.1.4 Other Roots

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Similar to the cube root, there is the fourth root, and the fifth root, and so on, as in the following examples:

$\begin{align*} \sqrt[4]{16}=2 \amp\text{ because }2^4=16\\ \sqrt[5]{-32}=-2 \amp\text{ because }(-2)^5=-32\\ \sqrt[6]{64}=2 \amp\text{ because }2^6=64\text{,} \end{align*}$

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To calculate the fifth root of

$-32$ with a calculator, try typing (-32)^(1/5) or root(5,-32).

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Definition 13.1.26.

The index of a radical is the number “ $\highlight{n}$ ” in $\sqrt[\highlight{n}]{\phantom{x}}\text{.}$ The symbol $\sqrt[n]{\phantom{x}}$ is read “the $n$ th root.” The plural of index is indices, as in “we can evaluate radicals of multiple indices in a single expression.”

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Fact 13.1.27. Domain of Radical Functions.

To find the domain of any even indexed radical function, set the radicand greater than or equal to zero. The solution set is the domain of the function.

The domain of any odd indexed radical of a polynomial is $(-\infty,\infty)\text{.}$

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Example 13.1.28.

Algebraically find the domain of the function $g$ where $g(x)=7-3\sqrt[4]{10-5x}\text{.}$
Algebraically find the domain of the function $h$ where $h(x)=4\sqrt[5]{2x-5}+1\text{.}$

Explanation

First, note that the index of this function is $4\text{.}$ By Fact 13.1.27, to find the domain of this function we must set the radicand greater or equal to zero and solve.

\begin{align*} 10-5x\amp\ge0\\ -5x\amp\ge\subtractright{10}\\ x\amp\mathbin{\highlight{\le}}\divideunder{-10}{-5}\\ x\amp\le2 \end{align*}

So, the domain of $g$ must be $(-\infty,2]\text{.}$
First, note that the index of the function $h$ for $h(x)=4\sqrt[5]{2x-5}+1$ is $5\text{.}$ By Fact 13.1.27, the domain is $(-\infty,\infty)\text{.}$

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Exercises 13.1.5 Exercises

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Review and Warmup

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1.

Evaluate the following.

$-\sqrt{4}={}$ .

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2.

Evaluate the following.

$-\sqrt{16}={}$ .

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3.

Evaluate the following.

$\sqrt{-25}=$ .

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4.

Evaluate the following.

$\sqrt{-36}=$ .

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5.

Evaluate the following.

$\displaystyle{\sqrt{{{\frac{36}{49}}}}={}}$ .

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6.

Evaluate the following.

$\displaystyle{\sqrt{{{\frac{49}{144}}}}={}}$ .

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7.

Without using a calculator, evaluate the expression.

$\displaystyle{\sqrt[3]{27} =}$
$\displaystyle{\sqrt[3]{-27} =}$
$\displaystyle{-\sqrt[3]{27} =}$

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8.

Without using a calculator, evaluate the expression.

$\displaystyle{\sqrt[4]{81} =}$
$\displaystyle{\sqrt[4]{-81} =}$
$\displaystyle{-\sqrt[4]{81} =}$

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9.

Without using a calculator, estimate the value of $\sqrt{98}\text{:}$

10.10
9.10
9.90
10.90

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10.

Without using a calculator, estimate the value of $\sqrt{17}\text{:}$

4.88
3.88
3.12
4.12

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11.

Without using a calculator, estimate the value of $\sqrt[3]{3}\text{:}$

0.56
0.44
1.44
1.56

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12.

Without using a calculator, estimate the value of $\sqrt[3]{5}\text{:}$

1.71
1.29
2.71
2.29

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