Q 60. - Precalculus Enhanced with Graphing Utilities

Question

The function $f\left(x\right)=\frac{4}{x+2}$ is one-to-one. Find its inverse and check the answer. Graph $f,f^{-1}$ and $y=x$ on the same coordinate axes.

Step-by-Step Solution

Answer

Inverse of the function $f\left(x\right)=\frac{4}{x+2}$ is $f^{-1}\left(x\right)=\frac{4-2x}{x}$. Graph is shown below.

Step 1. Given information.

Given a one-to-one function $f\left(x\right)=\frac{4}{x+2}$.

Step 2. Find the inverse of the function f and verify .

To find the explicit form of the inverse solve $x=\frac{4}{y+2}$ for y.

$\begin{array}{rcl}x& =& \frac{4}{y+2}\\ xy+2x& =& 4\\ xy& =& 4-2x\\ y& =& \frac{4-2x}{x}\end{array}$

It follows that $y=\frac{4-2x}{x}$.

Note that if $f\left(g\left(x\right)\right)=x$ and $g\left(f\left(x\right)\right)=x$, then f and g are inverse of one another.

Prove $f\left(f^{-1}\left(x\right)\right)=x$ as follows.

$\begin{array}{rcl}f\left(f^{-1}\left(x\right)\right)& =& f\left(\frac{4-2x}{x}\right)\\ & =& \frac{4}{\left(\frac{4-2x}{x}\right)+2}\\ & =& \frac{4x}{4-2x+2x}\\ & =& x\end{array}$

Prove $f^{-1}\left(f\left(x\right)\right)=x$ as follows.

$\begin{array}{rcl}{f}^{-1}\left(f\left(x\right)\right)& =& f^{-1}\left(\frac{4}{x+2}\right)\\ & =& \frac{4-2\left(\frac{4}{x+2}\right)}{\left(\frac{4}{x+2}\right)}\\ & =& \frac{4x+8-8}{4}\\ & =& x\end{array}$

Therefore, inverse of f is $f^{-1}\left(x\right)=\frac{4-2x}{x}$.

Step 3. Graph f , f - 1 and y = x on the same coordinate axes.

Graph $f\left(x\right)=\frac{4}{x+2},f^{-1}\left(x\right)=\frac{4-2x}{x}$ and $y=x$ as follows.

Note the symmetry of the graphs with respect to the line $y=x$.