Chapter 10.4, Problem 57E

To determine

To find the domain, range and inverse of the function.

Answer to Problem 57E

Change $x^2$ to $x^3$

Inverse of new function is $g\left(x\right)=\sqrt[3]{x+5}$

Explanation of Solution

Given information: $f(x)=x^2-5$

Formula Used: Horizontal line test and Inverse Relation

Calculation:

If a relation contains the point $\left(x,y\right)$ then the Inverse Relation contains the point $\left(y,x\right)$ because the inverse relation interchange the x and y coordinates.

  $f(x)=x^2-5$

As,

  $y=f(x)$

Now,

  $y=x^2-5$

  

This function does not passes the horizontal test.

So, this function does not have any inverse.

If change $x^2$ to $x^3$ then they have an inverse.

Now,

  $f(x)=x^3-5$

Domain = $\left(-\infty ,\infty \right)$

Range = $\left(-\infty ,\infty \right)$

As,

  $y=f(x)$

Now,

  $y=x^3-5$

  

This function passes the horizontal test.

So, the inverse is a function.

So,

  $x^3=y+5$

Now, cube root both sides.

  $\sqrt[3]{x^3}=\sqrt[3]{y+5}$

So,

  $x=\sqrt[3]{y+5}$

Hence,

Inverse of $f$ is $g\left(x\right)=\sqrt[3]{x+5}$