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The function, $f(x)=x \sqrt{1-x}$, where $x \in(0,1)$, has local maximum at $x=$
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Verified Answer
The correct answer is:
$\frac{2}{3}$
$\begin{aligned}
& f(x)=x \sqrt{1-x} \\
& \Rightarrow f^{\prime}(x)=1 \sqrt{1-x}+\frac{x}{2 \sqrt{1-x}} x(-1)=\frac{2-3 x}{2 \sqrt{1-x}}
\end{aligned}$
sign scheme of $f^{\prime}(x)$
Hence, local maxima at $x=\frac{2}{3}$
& 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 $x=\frac{2}{3}$
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