Question

If f(x)=x+5 and g(x)=$\sqrt{x^2-9}$ then find the domain of gof(x)

• A. (-8,-2)
• B. $(-\infty ,-8)\cup (-2,\infty )$
• C. $(-\infty ,-8]\cup [-2,\infty )$
• D. $(-(-\infty ,-8]\cup [-2,\infty )$

Answer 1

The correct option is C $(-\infty ,-8]\cup [-2,\infty )$

$gof\left(x\right)=g\left(f\right(x\left)\right)=\sqrt{{(x+5)}^2-9}=\sqrt{x^2+25+10x-9}$
$=\sqrt{x^2+16+10x}=\sqrt{(x+8)(x+2)}$
Now, for the square root to have a real value, both $(x+8)and(x+2)$ should be positive or equal to zero or both should be negative or equal to zero.
Case 1: When both
$x+8\ge 0$ and $x+2\ge 0$
$=>x\ge -8$ and $x\ge -2$
$=>x\ge -2$ is the solution for this case
Case 2: When both
$x+8\le 0$ and $x+2\le 0$
$=>x\le -8$ and $x\le -2$
$=>x\le -8$ is the solution for this case.
So, combining these both, we get the domain, $x\le -8andx\ge -2$