The domain of the function f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x

gmatphobia chetan2u Bunuel, how do we figure that we have to equate expressions as well?

That's because the even roots are defined for values that are more than or equal to zero.

We can PLUG IN THE ANSWERS, which represent the domain of the function.

Options A and B include x=0; the remaining options do not.
If x=0, we get:
\(f(0) = \sqrt{\sqrt{2} - \sqrt{4} } ≈ \sqrt{1.4 - 2} = \sqrt{-0.6} \)
Not viable, since the value under the root symbol may not be negative.
Since x=0 is not viable, eliminate A and B.

Options D and E include x=4; option C does not.
If x=4, we get:
\(f(4) = \sqrt{\sqrt{6} - \sqrt{0} } ≈ \sqrt{\sqrt{6}} \)
This works.
Eliminate C, which does not include x=4.

Options D includes x=1; option E does not.
If x=1, we get:
\(f(1) = \sqrt{\sqrt{3} - \sqrt{3} } = \sqrt{0} \)
This works.
Eliminate E, which does not include x=1.

I completely get the logics but can I understand from you why \(x \geq -2\), \(x \leq 4\), \(x \geq 1 \) at the end it is \(1 \leq x \leq 4\), why not  \(-2 \leq x \leq 4\)?
 

The conditions (\(x \geq -2\), \(x \leq 4\), \(x \geq 1\)) represent restrictive conditions, meaning for the whole expression to be defined they must be true simultaneously. You see, if \(-2 \leq x < 1\), then \(\sqrt{\sqrt{x + 2} - \sqrt{4 – x}} < 0\), making this expression undefined for real numbers in that range.