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प्रश्न
`f(x) = x/2+2/x, x>0 `.
उत्तर
\[\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{x}{2} + \frac{2}{x}\]
\[ \Rightarrow f'\left( x \right) = \frac{1}{2} - \frac{2}{x^2}\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \frac{1}{2} - \frac{2}{x^2} = 0\]
\[ \Rightarrow x^2 = 4\]
\[ \Rightarrow x = 2\text { and } - 2\]
\[\text { Thus, x = 2 and x = - 2 are the possible points of local maxima or a local minima } . \]
\[\text { Since }x > 0, x = 2\]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{4}{x^3}\]
\[\text { At }x = 2: \]
\[ f''\left( 2 \right) = \frac{4}{\left( 2 \right)^3} = \frac{1}{2} > 0\]
\[\text { So, x = 2 is the point of local minimum } . \]
\[\text { The local minimum value is given by } \]
\[f\left( 2 \right) = \frac{x}{2} + \frac{2}{x} = 1 + 1 = 2\]
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