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Section14.1Introduction to Radical Functions

We learned basics of square root in Section 1.3<<Unresolved xref, reference "section-square-root-properties.ptx"; check spelling or use "provisional" attribute>>, . In this chapter, we will learn roots of other powers and their applications.

Figure14.1.1Alternative Video Lesson

Subsection14.1.1The Square Root Function

We learned the definitions of radicals and radicands in Definition 8.2.2. If a function's variable is in a radicand, the function is called a radical function, such as \(f(x)=\sqrt{x}\text{.}\)

Scenario: An artist makes copper plates shaped in squares. Customers can choose the size. Build a function to calculate the length of a plate's side given its area. One customer ordered a plate with an area of \(6.25\) square feet. Calculate the length of this plate's side.

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

The function is \(l(A)=\sqrt{A}\text{.}\) We don't consider the negative solution in this context. For the sake of simplicity, we will replace \(l\) with \(f\text{,}\) and replace \(A\) with \(x\text{.}\) Let's build a table for the function \(f(x)=\sqrt{x}\text{:}\)

\(x\) \(f(x)\) Points on the Curve
\(0\) \(\sqrt{0}=0\) \((0,0)\)
\(1\) \(\sqrt{1}=1\) \((1,1)\)
\(4\) \(\sqrt{4}=2\) \((4,2)\)
\(6.25\) \(\sqrt{6.25}=2.5\) \((6.25,2.5)\)
\(9\) \(\sqrt{9}=3\) \((9,3)\)
Table14.1.2Values of \(f(x)=\sqrt{x}\)

We will plot points and look at the graph of \(f(x)=\sqrt{x}\text{.}\)

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.3Graph of \(f(x)=\sqrt{x}\)

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, both the function's domain and range are \([0,\infty)\text{.}\) The function's graph is missing in the Quadrant II and III, implying it is undefined in \((-\infty,0)\text{.}\) This makes sense because we cannot square root a negative number.

We will look at some examples on how to find the domain and range of square root functions.

Example14.1.4

Find the domain and range of \(g(x)=\sqrt{2x-4}+1\text{.}\)

Solution

Recall that the square root of a negative number is undefined. This implies the radicand of a square root must be greater or equal to zero. To find the function's domain, we set the radicand to be greater than or equal to zero:

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

The function's domain is \(x\ge2\text{,}\) or \([2,\infty)\) in interval notation.

To find the function's range, we use technology to look at the function's graph:

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.5Graph of \(g(x)=\sqrt{2x-4}+1\)

According to the graph, the function's range is \([1,\infty)\text{.}\) The graph also verifies the function's domain is indeed \([2,\infty)\text{.}\)

We could also find the function's range without graphing. We learned that the range of \(f(x)=\sqrt{x}\) is \([0,\infty)\text{.}\) This implies the range of \(\sqrt{2x-4}\) is \([0,\infty)\text{,}\) and thus the range of \(g(x)=\sqrt{2x-4}+1\) is \([1,\infty)\text{.}\)
Example14.1.6

Find the domain and range of \(h(x)=\frac{1}{\sqrt{4-2x}}\text{.}\)

Solution

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

\begin{align*} 4-2x\amp\ge0\\ -2x\amp\ge-4\\ \frac{-2x}{-2}\amp\le\frac{-4}{-2}\\ x\amp\le2 \end{align*}

When a function has a fraction, we must remember that dividing zero is not allowed. For this function, the denominator cannot be zero, which implies \(x\ne2\text{.}\) The function's domain is \((-\infty,2)\text{.}\)

To find the function's range, we use technology to look at the function's graph:

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.7Graph of \(h(x)=\frac{1}{\sqrt{4-2x}}\)

It looks like the function's value approaches \(0\) as \(x\)-value approaches \(-\infty\text{,}\) and the function's value approaches \(\infty\) as \(x\)-value approaches \(2\) from the left. We can verify this by looking at a table of \(h(x)\text{:}\)

\(x\) \(g=h(x)\)
\(-10000\) \(0.007\ldots\)
\(-20000\) \(0.004\ldots\)
\(-30000\) \(0.004\ldots\)
\(-40000\) \(0.003\ldots\)
\(x\) \(h(x)\)
\(1.99\) \(7.071\ldots\)
\(1.999\) \(22.36\ldots\)
\(1.9999\) \(70.71\ldots\)
\(1.99999\) \(223.6\dots\)

According to data in the table, our guesses were correct. The function's range is \((0,\infty)\text{.}\) Note that this function has a vertical asymptote at \(x=2\text{,}\) and a horizontal asymptote at \(y=0\text{.}\)

We will end this subsection with a real-life application of square root functions.

Example14.1.10

When an object is dropped with no initial velocity toward the ground, the time it takes to hit the ground, in seconds, can be modeled by the function

\begin{equation*} t(d)=\sqrt{\frac{2d}{9.8}} \end{equation*}

where \(d\) stands for the height of the object in meters. Answer the following questions with graphing technology.

  1. If an object is dropped from the top of a \(100\)-foot tall building, how long will it take for the object to hit the ground?

  2. \(3\) seconds since an object was released, assuming the object has not hit the ground yet, how far has the object traveled?

Solution

With graphing technology, and after adjusting the window settings, we can see the graph of the function \(t(d)=\sqrt{\frac{2d}{9.8}}\) and some critical points.

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.11Graph of \(t(d)=\sqrt{\frac{2d}{9.8}}\)

  1. The function passes the point \((100,4.52)\text{,}\) implying it will take the object approximately \(4.52\) seconds to hit the ground if it's released on top of a \(100\)-foot building.

  2. The function passes the point \((44.1,3)\text{,}\) implying the object has traveled approximately \(44.1\) meters \(3\) seconds since it was released.

Subsection14.1.2Exercises

Introduction to the Square Root Function

1
WeBWorK Exercise
2
WeBWorK Exercise
3
WeBWorK Exercise
4
WeBWorK Exercise
5
WeBWorK Exercise
6
WeBWorK Exercise
7
WeBWorK Exercise
8
WeBWorK Exercise

Subsection14.1.3The Distance Formula

Figure14.1.16Alternative Video Lesson

Square root is used in calculating the distance between two points on a coordinate plane. We learned the Pythagorean Theorem in Subsection 8.2.1. In a coordinate plane, we can use the Pythagorean Theorem to calculate the distance between two points.

Example14.1.17

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

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

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.18Calculating 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 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*}

Note that we didn't need to do \(d=\pm\sqrt{25}\) because distance must have positive values.

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

Example14.1.19

Calculate the distance between \((x_1,y_1)\) and \((x_2,y_2)\text{.}\) We use subscripts to identify the first pair \((x_1,y_1)\) and the second pair \((x_2,y_2)\text{.}\)

First, we will sketch a graph for 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.

<<SVG image is unavailable, or your browser cannot render it>>

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.20Calculating the Distance Between \((2,3)\) and \((5,7)\)
Figure14.1.21Calculating the Distance Between \((x_1,y_1)\) and \((x_2,y_2)\)

To calculate the distance between \((x_1,y_1)\) and \((x_2,y_2)\text{,}\) 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*}

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

Example14.1.22

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

Solution

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*}

With each value marked, it would be difficult to make the mistake of substituting a variable with the wrong value. We have:

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

Solution: The distance between \((-2,4)\) and \((5,-20)\) is \(25\) units.
Remark14.1.23

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 do

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

We should not do

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

Subsection14.1.4Exercises

Distance Formula

1
WeBWorK Exercise
2
WeBWorK Exercise
3
WeBWorK Exercise
4
WeBWorK Exercise

Subsection14.1.5Cube Root Function

The square of \(2\) is \(4\text{,}\) so the square root of \(4\) is \(2\text{.}\)

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*}

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

\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*}

One major difference between cube root and square root is that it's ok to cube root negative numbers. For example:

\begin{equation*} (-1)^3=-1,\quad\sqrt[3]{-1}=-1 \end{equation*}

However, we may not square root negative numbers. For example:

\begin{equation*} \sqrt{-1}=\text{undefined} \end{equation*}

Most calculators don't have the cube root button. When we calculate the cube root of a number, we can use the power button (which is usually marked with the wedge symbol, \(\wedge\)). For example, to calculate \(\sqrt[3]{8}\text{,}\) we press the following buttons:

\begin{equation*} 8\wedge(1/3) \end{equation*}

We will explain why \(\sqrt[3]{8}=8^{\frac{1}{3}}\) in Section 14.2. For now, just learn how to use a calculator to calculate the cube root of a given number.

We can also estimate the value of a cube root, like \(\sqrt[3]{10}\text{:}\)

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

By the values above, \(\sqrt[3]{10}\) must be between \(2\) and \(3\text{,}\) and closer to \(2\text{.}\) We can use a calculator to verify \(\sqrt[3]{10}=2.154...\)

We will build a table and graph the cube root function:

\(x\) \(g(x)=\sqrt[3]{x}\) Points on the Curve
\(-8\) \(\sqrt[3]{-8}=-2\) \((-8,-2)\)
\(-1\) \(\sqrt[3]{-1}=-1\) \((-1,-1)\)
\(0\) \(\sqrt[3]{0}=0\) \((0,0)\)
\(1\) \(\sqrt[3]{1}=1\) \((1,1)\)
\(8\) \(\sqrt[3]{8}=2\) \((8,2)\)
Table14.1.24Values of \(g(x)=\sqrt[3]{x}\)

We will plot points and look at the graph of \(g(x)=\sqrt[3]{x}\text{.}\)

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.25Graph of \(g(x)=\sqrt[3]{x}\)

Both the domain and range of \(g(x)=\sqrt[3]{x}\) are \((-\infty,\infty)\text{,}\) or simply "all real numbers." Compare this with the domain and range of \(f(x)=\sqrt{x}\text{,}\) which are both \([0,\infty)\text{.}\) The reason for the difference is that we may not square root negative numbers, but we may cube root negative numbers.

Remark14.1.26

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

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

The next example is an application of cube root function.

Example14.1.27

A factory makes customized plastic balls. A ball's volume can be calculated by the formula \(V(r)=\frac{4}{3}\pi r^3\text{,}\) where \(V\) stands for volume and \(r\) stands for radius. A customer asks for plastic balls with a volume of \(3\) cubic feet. Build a function to calculate the ball's radius, and then find the radius of those balls. Round your answer to two decimal places.

Solution

We will solve for \(r\) in the formula \(V(r)=\frac{4}{3}\pi r^3\text{:}\)

\begin{align*} V\amp=\frac{4}{3}\pi r^3\\ \multiplyleft{3}V\amp=\multiplyleft{3}\frac{4}{3}\pi r^3\\ 3V\amp=4\pi r^3\\ \frac{3V}{4\pi}\amp=\frac{4\pi r^3}{4\pi}\\ \frac{3V}{4\pi}\amp=r^3\\ \sqrt[3]{\frac{3V}{4\pi}}\amp=\sqrt[3]{r^3}\\ \sqrt[3]{\frac{3V}{4\pi}}\amp=r\\ r(V)\amp=\sqrt[3]{\frac{3V}{4\pi}} \end{align*}

Next, we substitute \(V\) in \(r(V)\) with \(3\text{,}\) and we have:

\begin{align*} r(V)\amp=\sqrt[3]{\frac{3V}{4\pi}}\\ r(3)\amp=\sqrt[3]{\frac{3(3)}{4\pi}}\\ r(3)\amp\approx0.89 \end{align*}

Solution: The plastic balls should have a radius of approximately \(0.89\) foot.

When we calculate \(\sqrt[3]{\frac{3(3)}{4\pi}}\) with a calculator, we press the following buttons:

\begin{equation*} (3*3/(4\pi))\wedge(1/3) \end{equation*}
Remark14.1.28

When we cube root both sides of an equation, like \(x^3=8\text{,}\) we do not need to add the \(\pm\) symbol like when we solve \(x^2=4\text{.}\) Compare the following equations:

\begin{align*} x^3\amp=8\\ \sqrt[3]{x^3}\amp=\sqrt[3]{8}\\ x\amp=2 \end{align*}

\begin{align*} x^2\amp=4\\ \sqrt{x^2}\amp=\pm\sqrt{4}\\ x\amp=\pm2 \end{align*}

For \(x^2=4\text{,}\) there are two solutions because \(2^2=4\) and \((-2)^2=4\text{.}\) However, for \(x^3=8\text{,}\) there is only one solution as \(2^3=8\text{.}\) The number \(-2\) is not a solution of \(x^3=8\text{,}\) because \((-2)^3=-8\text{.}\)

The following table summarized the difference of square root and cube root:

\(f(x)=\sqrt{x}\) \(g(x)=\sqrt[3]{x}\)
Root of a negative number \(\sqrt{-1}=\text{undefined}\) \(\sqrt[3]{-1}=-1\)
Solve \(x^n=1\)

\begin{align*} x^2\amp=1\\ x\amp=\pm1 \end{align*}

\begin{align*} x^3\amp=1\\ x\amp=1 \end{align*}
Solve \(x^n=-1\)

\begin{align*} x^2\amp=-1\\ x\amp\text{ is undefined} \end{align*}

\begin{align*} x^3\amp=-1\\ x\amp=-1 \end{align*}
Domain \([0,\infty)\) \((-\infty,\infty)\)
Range \([0,\infty)\) \((-\infty,\infty)\)
Table14.1.29Differences of \(f(x)=\sqrt{x}\) and \(g(x)=\sqrt[3]{x}\)

Subsection14.1.6Functions of Other Roots

Similar to the cube root, there are the \(4^{th}\) root, \(5^{th}\) root, etc, as in the following examples:

\begin{align*} \amp2^4=16,\amp\sqrt[4]{16}=2\\ \amp(-2)^5=-32,\amp\sqrt[5]{-32}=-2\\ \amp2^6=64,\amp\sqrt[6]{64}=2 \end{align*}

To calculate the \(5^{th}\) root of \(-32\) with a calculator, we press the following buttons:

\begin{equation*} (-32)\wedge(1/5) \end{equation*}

In the following table, we will summarize the differences between \(f(x)=x^m\) where \(m\) is even and \(g(x)=x^n\) where \(n\) is an odd number. The content of this table is very similar to content in Table 14.1.29.

\(f(x)=\sqrt[m]{x}\) where \(m\) is even \(g(x)=\sqrt[n]{x}\) where \(n\) is odd
Root of a negative number \(\sqrt[m]{-1}=\text{undefined}\) \(\sqrt[n]{-1}=-1\)
Solve \(x^p=1\)

\begin{align*} x^m\amp=1\\ x\amp=\pm1 \end{align*}

\begin{align*} x^n\amp=1\\ x\amp=1 \end{align*}
Solve \(x^p=-1\)

\begin{align*} x^m\amp=-1\\ x\amp\text{ is undefined} \end{align*}

\begin{align*} x^n\amp=-1\\ x\amp=-1 \end{align*}
Domain \([0,\infty)\) \((-\infty,\infty)\)
Range \([0,\infty)\) \((-\infty,\infty)\)
Table14.1.30Differences of \(f(x)=\sqrt[m]{x}\) where \(m\) is even and \(g(x)=\sqrt[n]{x}\) where \(n\) is odd

The graph of \(f(x)=\sqrt[m]{x}\text{,}\) where \(m\) is an even number, is very similar to the square root function's graph. They all pass the points \((0,0)\) and \((1,1)\text{.}\)

The graph of \(g(x)=\sqrt[n]{x}\text{,}\) where \(n\) is an odd number, is very similar to the cube root function's graph. They all pass the points \((-1,-1)\text{,}\) \((0,0)\) and \((1,1)\text{.}\)

Example14.1.31

Find the domain and range of \(g(x)=\sqrt[5]{2x-4}+1\text{.}\)

Solution

There is no limitation on the domain of a fifth-root function, so the domain is simply \((-\infty,\infty)\text{,}\) or "all real numbers."

To find the function's range, we use technology to graph the function:

<<SVG image is unavailable, or your browser cannot render it>>

Figure14.1.32Graph of \(g(x)=\sqrt[5]{2x-4}+1\)

According to the graph, the function's range is also "all real numbers."

Subsection14.1.7Exercises

Introduction to Radical Functions

1
WeBWorK Exercise
2
WeBWorK Exercise
3
WeBWorK Exercise
4
WeBWorK Exercise
5
WeBWorK Exercise
6
WeBWorK Exercise
7
WeBWorK Exercise
8
WeBWorK Exercise