If $f ( x ) = x \sqrt { 2 - x }$ , then $f ^ { \prime \prime } ( x ) = 0$ at x = 0 and x = 2.

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Quizplus · Accepted Answer

To find the second derivative $f''(x)$ of the function $f(x) = x\sqrt{2-x}$, we first find the first derivative $f'(x)$ and then differentiate it again. The given function is $f(x) = x\sqrt{2-x}$. Let's find its derivatives:1. First derivative $f'(x)$:$$f'(x) = \frac{d}{dx} [x(2-x)^{1/2}] = (2-x)^{1/2} + x\left(\frac{-1}{2}\right)(2-x)^{-1/2}(-1) = \sqrt{2-x} - \frac{x}{2\sqrt{2-x}}$$2. Second derivative $f''(x)$:$$f''(x) = \frac{d}{dx}\left[\sqrt{2-x} - \frac{x}{2\sqrt{2-x}}\right]$$This derivative involves applying the chain rule and the quotient rule, leading to a more complex expression. Without doing the full calculation here, it's clear that the second derivative will involve terms with $(2-x)^{-3/2}$ due to the differentiation of $\sqrt{2-x}$ and the quotient rule application. This means the second derivative will not simply equal zero at $x = 0$ and $x = 2$ without further specific conditions being met. The claim is false because the second derivative of the function does not automatically equal zero at these points without additional context or calculation.

If f(x)=x2−xf ( x ) = x \sqrt { 2 - x }f(x)=x2−x​

If $f ( x ) = x \sqrt { 2 - x }$