Chapter 6.3, Problem 22PPS

To determine

The domain and range of the function $f\left(x\right)=\sqrt{x+1}$ . And sketch the graph of the function.

Answer to Problem 22PPS

The domain of the function is $\left\{x|x\ge -1\right\}$ and range of the function is $\left\{y|y\ge 0\right\}$ .

Explanation of Solution

Given information: The given function is,

  $f\left(x\right)=\sqrt{x+1}\mathrm{....}\left(1\right)$

Calculation: The domain of the function is defined for which,

  $\begin{array}{c}x+1\ge 0\\ x\ge -1\end{array}$

Hence the domain of the function is $\left\{x|x\ge -1\right\}$ .

And range of the function can be evaluated by substituting the lower limit of domain in the function for which the function is defined. Therefore from (1),

  $\begin{array}{c}f\left(-1\right)=\sqrt{\left(-1\right)+1}\mathrm{......}\left(\text{For }x=-1\right)\\ =0\end{array}$

Hence the range of the function is $\left\{y|y\ge 0\right\}$

Graph: The graph for the function is shown as,