Chapter 7.2, Problem 23PPS

To determine

The inverse of the function $f\left(x\right)=5x^2$ and graph the function and its inverse.

Answer to Problem 23PPS

The inverse of function $f\left(x\right)=5x^2$ is $f^{-1}\left(x\right)=\pm \sqrt{\frac{x}{5}}$ .

Explanation of Solution

Given information: The given function is,

  $f\left(x\right)=5x^2$

Concept used: In order to find the inverse of a given function $f\left(x\right)$ write $f\left(x\right)=y$ and exchange the variables $\left(x\right)\text{ and }\left(y\right)$ and solve for the variable $\left(y\right)$ . At last write $y=f^{-1}\left(x\right)$ which is the required inverse of the function $f\left(x\right)$ .

Calculation: For the given function $f\left(x\right)=5x^2$ , the inverse is evaluated as,

  $\begin{array}{c}y=5x^2\mathrm{......}\left(\text{Plugging }f\left(x\right)=y\right)\\ \text{or, }\left(x\right)=5\left(y^2\right)\mathrm{......}\left(\text{Exchanging variables}\right)\\ y=\pm \sqrt{\frac{x}{5}}\mathrm{......}\left(\text{Solving for variable }\left(y\right)\right)\\ f^{-1}\left(x\right)=\pm \sqrt{\frac{x}{5}}\mathrm{......}\left(\text{Plugging }y=f^{-1}\left(x\right)\right)\end{array}$

Hence, the inverse of function $f\left(x\right)=5x^2$ is $f^{-1}\left(x\right)=\pm \sqrt{\frac{x}{5}}$ .

Graph: The graph for the function $f\left(x\right)=5x^2$ and its inverse function $f^{-1}\left(x\right)=\pm \sqrt{\frac{x}{5}}$ is shown as,