Q. 23 - Precalculus Enhanced with Graphing Utilities

Question

What is the domain of the function $\sqrt{x^2-16}$?

Step-by-Step Solution

Answer

The domain of the function is $\left\{x:x\le -4 \mathrm{or} x\ge 4\right\}$.

Step 1. Given information

The given function is $\sqrt{x^2-16}$.

Step 2.

• The square root function $\sqrt{x^2-16}$ is defined when $x^2-16\ge 0$.
• Consider style="max-width: none; vertical-align: -5px;" $g\left(x\right)=x^2-16$ and factor the expression.

$g\left(x\right)=(x-4)(x+4)$

• $g\left(x\right)=0$ implies $(x-4)=0$ or $(x+4)=0$.
• So, the x -intercepts of the curve are $(4,0),(-4,0)$.
• The value of $g\left(0\right)= -16$, so, the y-intercept is $(0,-16)$.
• Find the vertex of the curve as follows:

$\begin{array}{rcl}x& =& \frac{-b}{2a}\\ & =& 0\end{array}$

• So, the vertex is also at the y-intercept $(0,-16)$.

Step 3. Plot the function

• Plot the graph of the function by using the obtained intercepts and vertex.

Step 4. Find the region above x- axis

• From the graph, the region on and above the horizontal axis is $x\le -4 \mathrm{or} x\ge 4$.
• The inequality is of the greater than or equal to type, so the solution set is $x\le -4 \mathrm{or} x\ge 4$.