Question

Find the domain of this function:

f(x) = square root x^2 -16

Answer 1

\(\large f(x) = \sqrt{x^2-16}\)

the domain of that function, or that is, the values "x" can take safely
are the ones that will NOT make the radical amount, negative

Answer 2

Eh, didn't mean to ruin this for you =S deleted mine

Answer 3

hehe... didn't have to, a helping hand is always welcome

Answer 4

So should I post it or not? :S

Answer 5

\(\bf f(x) = \sqrt{x^2-16}\\ \quad \\
x = 4\qquad \sqrt{(4)^2-16}\implies\sqrt{16-16}\implies 0\\ \quad \\
x = -4\qquad \sqrt{(-4)^2-16}\implies\sqrt{16-16}\implies 0\\\quad \\
x = \pm 3\qquad \sqrt{(\pm 3)^2-16}\implies \sqrt{9-16}\implies\color{red}{\sqrt{-7}}\\ \quad \\
x = \pm 5\qquad \sqrt{(\pm 5)^2-16}\implies \sqrt{25-16}\implies \sqrt{9}\implies \pm3\)

so as you can see, 4 or -4 works well, no negatives
you go below that, you get a negative value
you go above that, you're ok

so that's the domain

Answer 6

@pitamar sure, post away

Answer 7

\[
f(x) = \sqrt{x^2 - 16} \\
x^2 - 16 \ge 0 \\
x^2 \ge 16 \\
|x| \ge 4 \\
x \ge 4 \text{ or } x \le -4
\]
Small fix also.. hehe..