Question

Which of the following is the domain of the function $f\left(x\right)=\frac{3-x}{\sqrt{x^2-9}}$?

• A. $-3>x>3$
• B. $-x\le x\le 3$
• C. $-3\le x<3$
• D. $x<-3$ or $x>3$
• E. $x\le -3$ or $x\ge 3$

Answer 1

The correct option is D $x<-3$ or $x>3$

For $f\left(x\right)$ to exist

$x^2-9\ne 0$ and $x^2-9\ge 0$

Combining we get $x^2-9>0$

$\Rightarrow |x|>3$

$\Rightarrow -3>x$ or $x>3$