Chapter 5.2, Problem 39AYU

In Problems 35-44, verify that the functions $f$ and $g$ are inverses of each other by showing that $f(g(x))=x$ and $g(f\left(x\right))=x$. Give any values of $x$ that need to be excluded from the domain of $f$ and the domain of $g$.

$f(x)=\frac{1}{x}$; $g(x)=\frac{1}{x}$

To determine

To find: Verify that the functions $f$ and $g$ are inverses of each other by showing that $f(g(x))=x$ and $g(f\left(x\right))=x$. Give any values of $x$ that need to be excluded from the domain of $f$ and the domain of $g$.

$f(x)=\frac{1}{x}$;

$g(x)=\frac{1}{x}$

Answer to Problem 39AYU

Solution:

The given functions $f$ and $g$ are verified.

The values of $x$ to be excluded from the domain of $f$ and the domain of $g$ are 0.

Explanation of Solution

Given:

By domain of a Composite Function,

The domain of a composite function $f\left(g\left(x\right)\right)$ is the set of those inputs $x$ in the domain of $g$ for which $g\left(x\right)$ is in the domain of $f$.

$g\left(x\right)=\frac{1}{x}$

Calculation:

The domain of $g$ is 0.

$f\left(x\right)=\frac{1}{x}$

The domain of $f$ is 0.

From those inputs, $x$, in the domain of $g$ for which $g\left(x\right)$ is in the domain of $f$. That is, exclude those inputs, $x$, from the domain of $g$ for which $g\left(x\right)$ is not in the domain of $f$. The resulting set is the domain of $f\circ g$. i.e., 0.

We know the composition of $f$ and $g$ is defined as.

$\left(f\cdot g\right)\left(x\right)=f\left(g\left(x\right)\right)$

$\left(f\cdot g\right)\left(x\right)=f\left(\frac{1}{x}\right)$

$\left(f\cdot g\right)\left(x\right)=\frac{1}{\frac{1}{x}}$

$\left(f\cdot g\right)\left(x\right)=x$

Also $\left(g.f\right)\left(x\right)=g\left(f\left(x\right)\right)$

$\left(g\cdot f\right)\left(x\right)=g\left(\frac{1}{x}\right)$

$\left(g\cdot f\right)\left(x\right)=\frac{1}{\frac{1}{x}}$

$\left(g\cdot f\right)\left(x\right)=x$