To solve this problem, we need to determine the values of [tex]\(x\)[/tex] for which the function [tex]\(f(x) = \sqrt{4 - x^2}\)[/tex] is defined as a real number.
The function involves a square root, so the expression inside the square root, [tex]\(4 - x^2\)[/tex], must be non-negative (i.e., [tex]\(4 - x^2 \geq 0\)[/tex]) for [tex]\(f(x)\)[/tex] to be defined as a real number.
Let's analyze the inequality [tex]\(4 - x^2 \geq 0\)[/tex]:
1. Start by isolating [tex]\(x^2\)[/tex]:
[tex]\[
4 - x^2 \geq 0
\][/tex]
2. Rearrange the inequality to:
[tex]\[
x^2 \leq 4
\][/tex]
3. Solve for the range of [tex]\(x\)[/tex]:
- This implies [tex]\(x^2 \leq 4\)[/tex]. Taking the square root of both sides, we get:
[tex]\[
-2 \leq x \leq 2
\][/tex]
Thus, the function [tex]\(f(x) = \sqrt{4 - x^2}\)[/tex] is defined for all values of [tex]\(x\)[/tex] in the interval [tex]\([-2, 2]\)[/tex].
Now, we need to determine for which of the given values [tex]\(f(x)\)[/tex] is NOT defined. The given choices are: [tex]\(-2\)[/tex], [tex]\(0\)[/tex], [tex]\(2\)[/tex], and [tex]\(4\)[/tex].
- [tex]\(x = -2\)[/tex]: [tex]\[
f(-2) = \sqrt{4 - (-2)^2} = \sqrt{4 - 4} = \sqrt{0} = 0 \quad \text{(defined)}
\][/tex]
- [tex]\(x = 0\)[/tex]: [tex]\[
f(0) = \sqrt{4 - 0^2} = \sqrt{4} = 2 \quad \text{(defined)}
\][/tex]
- [tex]\(x = 2\)[/tex]: [tex]\[
f(2) = \sqrt{4 - 2^2} = \sqrt{4 - 4} = \sqrt{0} = 0 \quad \text{(defined)}
\][/tex]
- [tex]\(x = 4\)[/tex]: [tex]\[
f(4) = \sqrt{4 - 4^2} = \sqrt{4 - 16} = \sqrt{-12} \quad \text{(NOT defined, as the square root of a negative number isn't a real number)}
\][/tex]
From this analysis, the function [tex]\(f(x) = \sqrt{4 - x^2}\)[/tex] is NOT defined for [tex]\(x = 4\)[/tex].
Therefore, the value of [tex]\(x\)[/tex] for which the function [tex]\(f(x) = \sqrt{4 - x^2}\)[/tex] is NOT defined as a real number is [tex]\(\boxed{4}\)[/tex].
