Chapter 10.4, Problem 38E

To determine

To check horizontal line test whether inverse of $f$ is a function and find inverse relation.

Answer to Problem 38E

Yes

  $f^{-1}\left(x\right)=\frac{16x^2-2}{5}$

Explanation of Solution

Given information: $f(x)=x^3$

Formula Used: Horizontal line test and inverse relation.

Calculation:

If the horizontal line intersects the graph of a function in all places at exactly one point, then the given function should have an inverse that is also a function. We say this function passes the horizontal line test.

  

As the horizontal line intersect with the graph of function at 1 point. So the function passes the horizontal test. So, the inverse of $f$ is a function.

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.

  $f(x)=-\frac{1}{4}\sqrt{5x+2}$

As,

  $y=f(x)$

So,

  $y=-\frac{1}{4}\sqrt{5x+2}$

Now, squaring both sides

  $y^2=\frac{1}{16}\times \left(5x+2\right)$

So, simplify

  $16y^2=5x+2$

Now,

  $x=\frac{16y^2-2}{5}$

Hence,

  $f^{-1}\left(x\right)=\frac{16x^2-2}{5}$