Question

Find the domain of $f\left(x\right)=\sqrt{(4-x{)}^2-5}$.

• A. $x\le -3$ or $x\ge 7$
• B. $2-\sqrt{5}\le x\le 2+\sqrt{5}$
• C. $x\le 4-\sqrt{5}$ or $x\ge 4+\sqrt{5}$
• D. $4-\sqrt{5}\le x\le 4+\sqrt{5}$
• E. $x\le 21$ or $x\ge 29$

Answer 1

The correct option is C $x\le 4-\sqrt{5}$ or $x\ge 4+\sqrt{5}$

Given, $f\left(x\right)=\sqrt{{(4-x)}^2-5}$

The term under square root must be $\ge 0$.

So, we get ${(4-x)}^2-5\ge 0$

$\Rightarrow {(4-x)}^2\ge 5$

$\Rightarrow (4-x)\ge \sqrt{5}$ or $(4-x)\le -\sqrt{5}$

$\Rightarrow x\le 4-\sqrt{5}$ or $x\ge 4+\sqrt{5}$