Q. 10 - Precalculus Enhanced with Graphing Utilities

Question

The function $f\left(x\right)=\frac{5}{x+2}$ is one to one. Find $f^{-1}$. Find the domain and the range of $f$ and the domain and the range of $f^{-1}$.

Step-by-Step Solution

Answer

$f^{-1}=\frac{5-2x}{x}$

The domain of $f$ is defined for all values of $x$ except for $x=-2$ and its range is, all the real numbers.

The domain of $f^{-1}$is defined for all values of $x$ except for $x=0$ and the range is all the real numbers.

Step 1 Given function is,

$f\left(x\right)=\frac{5}{x+2}$.

Find $f^{-1}$.

Solve the given function for $x$.

$$\begin{gathered} f\left(x\right)=\frac{5}{x+2} \\ f\left(x\right)\left(x+2\right)=5 \\ xf\left(x\right)+2f\left(x\right)=5 \\ xf\left(x\right)=5-2f\left(x\right) \\ x=\frac{5-2f\left(x\right)}{f\left(x\right)} \end{gathered}$$

Step 2 Now substitute f x = x , x = f - 1 in the obtained equation.

$f^{-1}=\frac{5-2x}{x}$

The domain of $f\left(x\right)$ is defined for all values of $x$ except for $x=-2$ and the given function can take up any of the values in the range of real numbers . So, the range of $f$ is all the real numbers.

The domain of $f^{-1}$ is defined for all the values of $x$ except for $x=0$ and the given function can take up any of the values in the range of real numbers . So, the range of $f^{-1}$ is all the real numbers.