Find the critical points of the given function.
$f(x)=\(\sqrt{2-x^2}\)$

• A. $x=0$
• B. $x=0,x=\sqrt{2},x=-\sqrt{2}$
• C. $x=0,x=\(\sqrt\)2$
• D. No critical points

Answer: (B)

Verified step by step guidance

1. First, understand that critical points occur where the derivative of the function is zero or undefined. We need to find the derivative of the function f(x) = √(2 - x^2).
2. To find the derivative, apply the chain rule. The outer function is the square root, and the inner function is (2 - x^2). The derivative of the square root function √u is (1/2√u) * du/dx.
3. Calculate the derivative of the inner function (2 - x^2), which is -2x. Now, apply the chain rule: f'(x) = (1/2√(2 - x^2)) * (-2x).
4. Simplify the expression for the derivative: f'(x) = -x / √(2 - x^2).
5. Set the derivative equal to zero to find where it is zero: -x / √(2 - x^2) = 0. Solve for x to find the critical points. Additionally, check where the derivative is undefined, which occurs when the denominator is zero, i.e., when 2 - x^2 = 0.