What is the domain and range of $g (x) = {(x^{2} - 16)}^{\frac{1}{2}}$ $g(x)= (x^2 - 16)^(1/2)$ ?

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Answer 1 · 2018-06-09T12:44:00

$x \in (- \infty, - 4] \cup [4, \infty)$ $x\in(-\infty, -4]\uu [4, \infty)$
$y \in [0, \infty]$ $y\in[0,\infty]$

$g (x) = \sqrt{x^{2} - 16}$ $g(x) = \sqrt{x^2-16}$

Anything underneath a square root must be greater than zero for the square root to be defined for real numbers.

$\therefore x^2-16>0 \Rightarrow x^2>16 \Rightarrow |x|>4$

$\therefore x\in(-\infty, -4]\uu [4, \infty)$

The range will be $[0,\infty)$ since the expression inside the sqrt ranges from 0 to $\infty$

$\therefore y\in[0,\infty]$

What is the domain and range of g(x)= (x^2 - 16)^(1/2)g(x)=(x2−16)12g(x)= (x^2 - 16)^(1/2)?