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ID:
Class: 12
Subject: Math
Topic: Application of Derivative
Type: Short
Question:
Find the local maxima and minima of \( f(x) = x \sqrt{1 - x}, \, x > 0 \).
Official Solution
Explanation:
Let us differentiate:
\[
f(x) = x \sqrt{1 - x} \Rightarrow f'(x) = \sqrt{1 - x} + x \cdot \left( \frac{-1}{2\sqrt{1 - x}} \right)
= \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}}
\]
Set \( f'(x) = 0 \Rightarrow 2 - 3x = 0 \Rightarrow x = \frac{2}{3} \)
Now second derivative:
\[
f''(x) = \frac{d}{dx} \left( \frac{2 - 3x}{2\sqrt{1 - x}} \right)
= \frac{3x - 4}{4(1 - x)^{3/2}}
\Rightarrow f''\left( \frac{2}{3} \right) = \frac{2 - 4}{2(1/3)^{3/2}} < 0
\Rightarrow \text{Local maximum}
\]
\[
f\left( \frac{2}{3} \right) = \frac{2}{3} \cdot \sqrt{\frac{1}{3}} = \frac{2\sqrt{3}}{9}
\]
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