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प्रश्न
`f(x)=xsqrt(32-x^2), -5<=x<=5` .
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उत्तर
\[\text { Given }: f\left( x \right) = x\sqrt{32 - x^2}\]
\[ \Rightarrow f'\left( x \right) = \sqrt{32 - x^2} - \frac{x^2}{\sqrt{32 - x^2}}\]
\[\text {For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \sqrt{32 - x^2} - \frac{x^2}{\sqrt{32 - x^2}} = 0\]
\[ \Rightarrow \sqrt{32 - x^2} = \frac{x}{\sqrt{32 - x^2}}\]
\[ \Rightarrow 32 - x^2 = x^2 \]
\[ \Rightarrow x^2 = 16\]
\[ \Rightarrow x = \pm 4 \]
\[\text { Thus, x = 4 and x = - 4 are the possible points of local maxima or local minima }. \]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{- x}{\sqrt{32 - x^2}} - \left( \frac{2x\sqrt{32 - x^2} + \frac{x^3}{\sqrt{32 - x^2}}}{32 - x^2} \right) = \frac{- x}{\sqrt{32 - x^2}} - \left( \frac{2x\left( 32 - x^2 \right) + x^3}{\left( 32 - x^2 \right)\sqrt{32 - x^2}} \right)\]
\[\text { At }x = 4: \]
\[ f''\left( 4 \right) = \frac{- 4}{\sqrt{32 - 4^2}} - \left[ \frac{8\left( 32 - 4^2 \right) + 4^3}{\left( 32 - 4^2 \right)\sqrt{32 - 4^2}} \right] = - 1 - \frac{192}{64} = - 3 < 0\]
\[\text { So, x = 4 is the point of local maximum } . \]
\[\text { The local maximum value is given by} \]
\[f\left( 4 \right) = 4\sqrt{32 - 4^2} = 16\]
\[\text { At } x = - 4: \]
\[ f''\left( - 4 \right) = \frac{4}{\sqrt{32 - 4^2}} + \left[ \frac{8\left( 32 - 4^2 \right) - 4^3}{\left( 32 - 4^2 \right)\sqrt{32 - 4^2}} \right] = 1 + 2 = 3 > 0\]
\[\text { So, x = - 4 is the point of local minimum } . \]
\[\text { The local minimum value is given by } \]
\[f\left( - 4 \right) = - 4\sqrt{32 - 4^2} = - 16\]
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