Chapter A, Problem 1E

In Exercises l-6, show that f and g are inverses of each other by showing that f[g(x)] = x and g[f(x)] = x.

1. $f(x)=\frac{1}{3}{x}^3;g(x)=\sqrt[3]{3x}$

To determine

To show: The function f and g are inverse of each other.

Explanation of Solution

Given:

The given functions are $f\left(x\right)=\frac{1}{3}{x}^3$, $g\left(x\right)=\sqrt[3]{3x}$.

Proof:

The domain and rang of the both functions are $\left(-\infty ,\infty \right)$.

Check the composition $f\left[g\left(x\right)\right]=x$.

$\begin{array}{c}f\left[g\left(x\right)\right]=f\left[\sqrt[3]{3x}\right]\\ =\frac{1}{3}{\left(\sqrt[3]{3x}\right)}^3\\ =\frac{1}{3}{\left(3x\right)}^{3\cdot \frac{1}{3}}\\ =\frac{1}{3}\left(3x\right)\end{array}$

That is, $f\left[g\left(x\right)\right]=x$.

Check the composition $g\left[f\left(x\right)\right]=x$.

$\begin{array}{c}g\left[f\left(x\right)\right]=g\left[\left(\frac{1}{3}{x}^3\right)\right]\\ =\sqrt[3]{3\left(\frac{1}{3}{x}^3\right)}\\ =\sqrt[3]{x^3}\\ =x\end{array}$

That is, $g\left[f\left(x\right)\right]=x$.

Hence, the two functions f and g are inverse of each other is showed.