The function f(x) = xsqrt{1 - x^2} for x 0 | vedclass

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(A) Let $f(x) = x\sqrt{1 - x^2}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 \cdot \sqrt{1 - x^2} + x \cdot \fr

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$A$ local maxima

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(A) Let $f(x) = x\sqrt{1 - x^2}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 \cdot \sqrt{1 - x^2} + x \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x)$$f'(x) = \sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}} = \frac{1 - x^2 - x^2}{\sqrt{1 - x^2}} = \frac{1 - 2x^2}{\sqrt{1 - x^2}}$.Setting $f'(x) = 0$,we get $1 - 2x^2 = 0$,which implies $x^2 = \frac{1}{2}$,so $x = \pm \frac{1}{\sqrt{2}}$.Since the domain is $x > 0$,we consider $x = \frac{1}{\sqrt{2}}$.Now,we use the second derivative test:$f''(x) = \frac{d}{dx} \left( \frac{1 - 2x^2}{\sqrt{1 - x^2}} \right) = \frac{(-4x)(\sqrt{1 - x^2}) - (1 - 2x^2)(\frac{-x}{\sqrt{1 - x^2}})}{1 - x^2}$$f''(x) = \frac{-4x(1 - x^2) + x(1 - 2x^2)}{(1 - x^2)^{3/2}} = \frac{-4x + 4x^3 + x - 2x^3}{(1 - x^2)^{3/2}} = \frac{2x^3 - 3x}{(1 - x^2)^{3/2}}$.Evaluating at $x = \frac{1}{\sqrt{2}}$:$f''\left(\frac{1}{\sqrt{2}}\right) = \frac{2(\frac{1}{2\sqrt{2}}) - 3(\frac{1}{\sqrt{2}})}{(1 - 1/2)^{3/2}} = \frac{\frac{1}{\sqrt{2}} - \frac{3}{\sqrt{2}}}{(1/2)^{3/2}} = \frac{-2/\sqrt{2}}{(1/2)^{3/2}} Since $f''\left(\frac{1}{\sqrt{2}}\right) < 0$,the function has a local maxima at $x = \frac{1}{\sqrt{2}}$.

The function $f(x) = x\sqrt{1 - x^2}$ for $x > 0$ has:

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