Chapter 6.3, Problem 1BGP

To determine

To find: The domain and the range of given function.

Answer to Problem 1BGP

The domain is $\left\{x:x\ge -6\right\}$ and the range is $\left\{y:y\ge 2\right\}$ .

Explanation of Solution

Given:

The given function is:

  $f\left(x\right)=\sqrt{x+6}+2\dots \left(1\right)$

The value of what’s under the square root must be greater than equal to zero.

Therefore,

  $\begin{array}{c}x+6\ge 0\\ x\ge -6\end{array}$

So the domain is $\left\{x:x\ge -6\right\}$ .

When,

  $x=-6$ ,

Further simplified as:

  $\begin{array}{c}f\left(-6\right)=\sqrt{-6+6}+2\\ =2\end{array}$ .

And

  $\begin{array}{c}\underset{x\to +\infty }{\lim }f\left(x\right)=\underset{x\to +\infty }{\lim }\sqrt{x+6}+2\\ =+\infty \end{array}$

Therefore the range is $\left\{y:y\ge 2\right\}$ .