Chapter EP, Problem 9.3.3EP

To determine

To state: The domain and range of the given function and draw the graph of the function

Answer to Problem 9.3.3EP

Domain and range of the function $f\left(x\right)=\frac{1}{x+2}$ are $ℝ-\left\{-2\right\}$ and $ℝ$ respectively

Explanation of Solution

Given: $f\left(x\right)=\frac{1}{x+2}$

The given $f\left(x\right)=\frac{1}{x+2}$ and graph of the function $f\left(x\right)=\frac{1}{x+2}$ is

Now it has to state domain and range of the given function $f\left(x\right)=\frac{1}{x+2}$

Domain of the function $f\left(x\right)=\frac{1}{x+2}$ is $ℝ-\left\{-2\right\}$ .

  $\begin{array}{l}f\left(x\right)=\frac{1}{x+2}\\ f\left(-2\right)=\frac{1}{-2+2}\\ =\frac{1}{0}\\ =\infty \end{array}$

Range of the function $f\left(x\right)=\frac{1}{x+2}$ is $ℝ$ .

Hence, domain and range of the function $f\left(x\right)=\frac{1}{x+2}$ are $ℝ-\left\{-2\right\}$ and $ℝ$ respectively.