Chapter 4.3, Problem 7QR

To determine

To Find: The inverse of the given function.

Answer to Problem 7QR

The inverse is $f^{-1}(x)=x^3-5$

Explanation of Solution

Given information: Given function $y=\sqrt[3]{x+5}$

Calculation:

First we have to prove that the given function $f(x)=\sqrt[3]{x+5}$ is one-one onto(bijective).

For one-one, we have

  $\begin{array}{l}f(x_1)=f(x_2)\\ \sqrt[3]{x_1+5}=\sqrt[3]{x_2+5}\\ x_1+5=x_2+5\\ x_1=x_2\\ \therefore f\text{is}\text{one-one}\text{.}\end{array}$

For onto, we have

  $\begin{array}{l}y=\sqrt[3]{x+5}\\ y^3=x+5\\ x=y^3-5.\end{array}$

Now, if $y\in R$ , then there exists $x=y^3-5\in R$ such that

  $\begin{array}{l}f(x)=f\left(y^3-5\right)=\sqrt[3]{(y^3-5)+5}=y\\ \therefore f\text{is}\text{onto}\text{.}\end{array}$

Thus, $f$ is bijective and therefore invertible.

Now,

  $\begin{array}{l}f\circ f^{-1}(x)=x\\ f(f^{-1}(x))=x\\ \sqrt[3]{f^{-1}(x)+5}=x\\ \Rightarrow f^{-1}(x)=x^3-5\end{array}$