8.7 Use Radicals in Functions

ⓐ
$\begin{array}{cccccc}& & & \hfill f\left(x\right)& =\hfill & \sqrt{2x-1}\hfill \\ \text{To evaluate}f\left(5\right),\text{substitute 5 for}x.\hfill & & & \hfill f\left(5\right)& =\hfill & \sqrt{2·5-1}\hfill \\ \text{Simplify.}\hfill & & & \hfill f\left(5\right)& =\hfill & \sqrt{9}\hfill \\ \text{Take the square root.}\hfill & & & \hfill f\left(5\right)& =\hfill & 3\hfill \end{array}$

ⓑ
$\begin{array}{cccccc}& & & \hfill f\left(x\right)& =\hfill & \sqrt{2x-1}\hfill \\ \text{To evaluate}f\left(\mathrm{-2}\right),\text{substitute}\mathrm{-2}\text{for}x.\hfill & & & \hfill f\left(\mathrm{-2}\right)& =\hfill & \sqrt{2\left(\mathrm{-2}\right)-1}\hfill \\ \text{Simplify.}\hfill & & & \hfill f\left(\mathrm{-2}\right)& =\hfill & \sqrt{\mathrm{-5}}\hfill \end{array}$

Since the square root of a negative number is not a real number, the function does not have a value at $x=\mathrm{-2}.$

ⓐ
$\begin{array}{cccccc}& & & \hfill g\left(x\right)& =\hfill & \sqrt[3]{x-6}\hfill \\ \text{To evaluate}g\left(14\right),\text{substitute 14 for}x.\hfill & & & \hfill g\left(14\right)& =\hfill & \sqrt[3]{14-6}\hfill \\ \text{Simplify.}\hfill & & & \hfill g\left(14\right)& =\hfill & \sqrt[3]{8}\hfill \\ \text{Take the cube root.}\hfill & & & \hfill g\left(14\right)& =\hfill & 2\hfill \end{array}$

ⓑ
$\begin{array}{cccccc}& & & \hfill g\left(x\right)& =\hfill & \sqrt[3]{x-6}\hfill \\ \text{To evaluate}g\left(\mathrm{-2}\right),\text{substitute}\mathrm{-2}\text{for}x.\hfill & & & \hfill g\left(\mathrm{-2}\right)& =\hfill & \sqrt[3]{\mathrm{-2}-6}\hfill \\ \text{Simplify.}\hfill & & & \hfill g\left(\mathrm{-2}\right)& =\hfill & \sqrt[3]{\mathrm{-8}}\hfill \\ \text{Take the cube root.}\hfill & & & \hfill g\left(\mathrm{-2}\right)& =\hfill & \mathrm{-2}\hfill \end{array}$

ⓐ
$\begin{array}{cccccc}& & & \hfill f\left(x\right)& =\hfill & \sqrt[4]{5x-4}\hfill \\ \text{To evaluate}f\left(4\right),\text{substitute 4 for}x.\hfill & & & \hfill f\left(4\right)& =\hfill & \sqrt[4]{5·4-4}\hfill \\ \text{Simplify.}\hfill & & & \hfill f\left(4\right)& =\hfill & \sqrt[4]{16}\hfill \\ \text{Take the fourth root.}\hfill & & & \hfill f\left(4\right)& =\hfill & 2\hfill \end{array}$

ⓑ
$\begin{array}{cccccc}& & & \hfill f\left(x\right)& =\hfill & \sqrt[4]{5x-4}\hfill \\ \text{To evaluate}f\left(\mathrm{-12}\right),\text{substitute}\mathrm{-12}\text{for}x.\hfill & & & \hfill f\left(\mathrm{-12}\right)& =\hfill & \sqrt[4]{5\left(\mathrm{-12}\right)-4}\hfill \\ \text{Simplify.}\hfill & & & \hfill f\left(\mathrm{-12}\right)& =\hfill & \sqrt[4]{\mathrm{-64}}\hfill \end{array}$

Since the fourth root of a negative number is not a real number, the function does not have a value at $x=\mathrm{-12}.$

Since the function, $f\left(x\right)=\sqrt{3x-4}$ has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0. We set the radicand to be greater than or equal to 0 and then solve to find the domain.

$\begin{array}{cccccccc}& & & & & \hfill 3x-4& \ge \hfill & 0\hfill \\ \text{Solve.}\hfill & & & & & \hfill 3x& \ge \hfill & 4\hfill \\ & & & & & \hfill x& \ge \hfill & \frac{4}{3}\hfill \end{array}$

The domain of $f\left(x\right)=\sqrt{3x-4}$ is all values $x\ge \frac{4}{3}$ and we write it in interval notation as $\left[\frac{4}{3},\infty \right).$

Since the function, $g\left(x\right)=\sqrt{\frac{6}{x-1}}$ has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0.

The radicand cannot be zero since the numerator is not zero.

For $\frac{6}{x-1}$ to be greater than zero, the denominator must be positive since the numerator is positive. We know a positive divided by a positive is positive.

We set $x-1>0$ and solve.

$\begin{array}{cccccccc}& & & & & \hfill x-1& >\hfill & 0\hfill \\ \text{Solve.}\hfill & & & & & \hfill x& >\hfill & 1\hfill \end{array}$

Also, since the radicand is a fraction, we must realize that the denominator cannot be zero.

We solve $x-1=0$ to find the value that must be eliminated from the domain.

$\begin{array}{cccccccc}& & & & & \hfill x-1& =\hfill & 0\hfill \\ \text{Solve.}\hfill & & & & & \hfill x& =\hfill & 1\text{so}x\ne 1\text{in the domain.}\hfill \end{array}$

Putting this together we get the domain is $x>1$ and we write it as $\left(1,\infty \right).$

Since the function, $f\left(x\right)=\sqrt[3]{2x^2+3}$ has a radical with an index of 3, which is odd, we know the radicand can be any real number. This tells us the domain is any real number. In interval notation, we write $\left(\text{−}\infty ,\infty \right).$

The domain of $f\left(x\right)=\sqrt[3]{2x^2+3}$ is all real numbers and we write it in interval notation as $\left(\text{−}\infty ,\infty \right).$

ⓐ Since the radical has index 2, we know the radicand must be greater than or equal to zero. If $x+3\ge 0,$ then $x\ge \mathrm{-3}.$ This tells us the domain is all values $x\ge \mathrm{-3}$ and written in interval notation as $\left[\mathrm{-3},\infty \right).$

ⓑ To graph the function, we choose points in the interval $\left[\mathrm{-3},\infty \right)$ that will also give us a radicand which will be easy to take the square root.

ⓒ Looking at the graph, we see the y-values of the function are greater than or equal to zero. The range then is $\left[0,\infty \right).$

ⓐ Since the radical has index 3, we know the radicand can be any real number. This tells us the domain is all real numbers and written in interval notation as $\left(\text{−}\infty ,\infty \right)$

ⓑ To graph the function, we choose points in the interval $\left(\text{−}\infty ,\infty \right)$ that will also give us a radicand which will be easy to take the cube root.

ⓒ Looking at the graph, we see the y-values of the function are all real numbers. The range then is $\left(\text{−}\infty ,\infty \right).$