Chapter 6.3, Problem 16PPS

To determine

The domain and range of the function $f\left(x\right)=\sqrt{x+2}+5$ .

Answer to Problem 16PPS

The domain of the function is $\left\{x|x\ge -2\right\}$ and range of the function is $\left\{y|y\ge 5\right\}$ .

Explanation of Solution

Given information: The given function is,

  $f\left(x\right)=\sqrt{x+2}+\mathrm{5......}\left(1\right)$

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

  $\begin{array}{c}x+2\ge 0\\ x\ge -2\end{array}$

Hence the domain of the function is $\left\{x|x\ge -2\right\}$ .

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),

  $\begin{array}{c}f\left(-2\right)=\sqrt{\left(-2\right)+2}+\mathrm{5......}\left(\text{For }x=-2\right)\\ =5\end{array}$

Hence the range of the function is $\left\{y|y\ge 5\right\}$