The function,f(x)=x sqrt{1-x},where x in(0,1) | vedclass

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(C) Given function: $f(x) = x \sqrt{1-x}$To find the local maximum,we first find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = (1) \cdot \sqrt{1-x}

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$\frac{2}{3}$

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(C) Given function: $f(x) = x \sqrt{1-x}$To find the local maximum,we first find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = (1) \cdot \sqrt{1-x} + x \cdot \frac{1}{2\sqrt{1-x}} \cdot (-1)$$f^{\prime}(x) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}}$$f^{\prime}(x) = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$Set $f^{\prime}(x) = 0$ to find critical points:$\frac{2 - 3x}{2\sqrt{1-x}} = 0 \Rightarrow 2 - 3x = 0 \Rightarrow x = \frac{2}{3}$Now,check the sign of $f^{\prime}(x)$ around $x = \frac{2}{3}$:For $x  0$ (function is increasing).For $x > \frac{2}{3}$,$f^{\prime}(x) Since the derivative changes sign from positive to negative at $x = \frac{2}{3}$,the function has a local maximum at $x = \frac{2}{3}$.

The function,$f(x)=x \sqrt{1-x}$,where $x \in(0,1)$,has a local maximum at $x=$

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