Question

The domain of function $f\left(x\right)=\sqrt{-4\sqrt{5+2x}}$ for real $x$ is

• A. $\frac{3}{4}$
• B. $\frac{-3}{4}$
• C. $\frac{-5}{2}$
• D. $\frac{5}{2}$

Answer 1

Answer: (C)

$\frac{-5}{2}$

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

For function to be defined, the term under the square root should be greater than or equal to zero.

that is,

$-4\sqrt{5+2x}\ge 0$

$\sqrt{5+2x}⩽0$

but $\sqrt{5+2x}\ge 0$ because it is a square-root function.

So, for the function to be defined,

$5+2x=0$

$x=-\frac{5}{2}$