How do you find the domain of f(x)=sqrt((x^2-16))?

Apr 26, 2017

$x \ge 4 \mathmr{and} x \le - 4$

Explanation:

Domain for a function means the value/s of $x$ for which the function is valid. In this case, the function is invalid if $\left({x}^{2} - 16\right)$ is $<$ 0 as there are no real values of the function. So, you find the values of $x$ which would make $\left({x}^{2} - 16\right)$ greater or equal to 0 as the function would be valid.
So,

$\left({x}^{2} - 16\right) \ge 0$
${x}^{2} \ge 16$
${x}^{2} \ge {4}^{2}$
$| x | \ge 4$
Either,

$x \ge 4$

Or,

$x \le - 4$

So for the values of $x \ge 4 \mathmr{and} x \le - 4$, $\left({x}^{2} - 16\right)$ will be greater or equal to zero which will be give some value under square root making the function valid.
The remaining values $- 4 < x < 4$, would give you a negative term for $\left({x}^{2} - 16\right)$ which would not be valid under square root making the function invalid.

SO the domain is $x \ge 4 \mathmr{and} x \le - 4$