Chapter 6.7, Problem 45PPSE

Solution

To find: The inverse of the function $f\left(x\right)=1.5x^2-4$ and state if the inverse is a function or not.

The inverse of the function is $f^{-1}\left(x\right)=\pm \sqrt{\frac{2\left(x+4\right)}{3}}$ and it’s not a function.

Given information:

The function, $f\left(x\right)=1.5x^2-4$ .

Concept used:

Inverse $f^{-1}\left(x\right)$ of a function $f\left(x\right)$ is a function if for every $x$ in the domain of $f^{-1}\left(x\right)$ there is only one $y$ value in the range of $f^{-1}\left(x\right)$ .

Calculation:

Find the inverse of the function $f\left(x\right)=1.5x^2-4$ .

  $\begin{array}{c}f\left(x\right)=1.5x^2-4\\ y=1.5x^2-4\text{Rewrite the function with }y\\ x=1.5y^2-4\text{Switch }x\text{ and }y\\ 1.5y^2-4=x\text{Solve for }y\\ 1.5y^2=x+4\\ 15y^2=10x+40\\ y^2=\frac{2\left(x+4\right)}{3}\\ y=\pm \sqrt{\frac{2\left(x+4\right)}{3}}\end{array}$

So, $f^{-1}\left(x\right)=\pm \sqrt{\frac{2\left(x+4\right)}{3}}$ .

The inverse is not a function since for every $x$ in the domain of $f^{-1}\left(x\right)$ there are two $y$ values in the range of $f^{-1}\left(x\right)$ .

Conclusion:

The inverse of the function is $f^{-1}\left(x\right)=\pm \sqrt{\frac{2\left(x+4\right)}{3}}$ and it’s not a function.