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

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=x \sqrt{1-x} \ & \Rightarrow f^{\prime}(x)=1 \sqrt{1-x}+\frac{x}{2 \sqrt{1-x}} x(-1)=\frac{2-3 x}{2 \sqrt{1-x}} \end{aligned}$ sign scheme of $f^{\prime}(x)$ Hence, local maxima at $x=\frac{2}{3}$

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

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