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 (\(x \geq -2\), \(x \leq 4\), and \(x \geq 1\)) 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.­

would plugging in the values help here?

­difficult to get roots particularly ­combined roots with plugging in and time constraints
\(\sqrt{x+2} - \sqrt{4-x}\)
­

jefftargettesyprep I understand first part, can you please elaborate the second part

Step 1: outer square root to be defined
√(x+2) ≥ √(4−x)
x+2 ≥ 4−x
2x ≥ 2
X ≥ 1

Step 2: inner square roots to be defined
√(x+2) exists when: x≥−2
√(4−x) exists when:x≤4x

Combine all conditions:
From Step 1: x≥1
From Step 2: −2≤x≤4

Answer: 1≤x≤4

@[color=#000000]JeffTargetTestPrep[/color]
When you equated the two square roots in the last stage shouldnt you have first equated the mod and then squared the mod? How could you assume the same sign of inequality by skipping that step? Is that allowed? Bunuel

Solved the question using options.

Here is the detailed explanation



Smart Prep (Tutor)

Responding to a pm:

Domain means the values of x for which the expression will take real values. For example, we cannot put x = -5 because then (x+2), which is under a square root, will become negative. That is not allowed since we cannot find the square root of negative numbers.
Usually, when looking for domain, the common constraints are that the square root should not be negative and the denominator should not be 0.

There are multiple ways of solving this question.

Method 1: Ensure all square roots are non negative so we need to ensure that x + 2 >= 0, 4 - x >= 0 and \(\sqrt{x+2} - \sqrt{4-x} \geq 0\)
Show by solutions above.

Method 2: Plug in Values
I would prefer this since it is very quick. Simplest would be to check for x = 0, 3 and 1
Every value of x for which f(x) has a real value is a part of the domain.

When we put x = 0, we see that f(x) is not real and hence option (A) and (B) cannot be the domain since as per them, x can take value 0.
When we put x = 3, we see that f(x) is real so x = 3 is possible. So eliminate option (C) because it does not include 3.
When we put x = 1, we see that f(x) is real so eliminate option (E) because it does not include 1.

Answer (D)

How do we know that the numbers under the square roots must >= 0? The question says that the domain of x is a set of real numbers, but does not say the range will be real (i.e. that numbers under the square root must be positive & squart root value must be real)?

GMAT does not test on imaginary or complex numbers, hence it is fair to assume that values underneath the square root are greater than or equal to 0.

Also think, if the problem allowed complex number in range couldn't domain include any value of x, even those not mentioned in the answer choices.