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प्रश्न
Write the maximum value of f(x) = \[x + \frac{1}{x}, x > 0 .\]
योग
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उत्तर
\[\text { Given: } \hspace{0.167em} f\left( x \right) = x + \frac{1}{x}\]
\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]
\[\text { For a local maxima or a local minima, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]
\[ \Rightarrow x^2 = 1\]
\[ \Rightarrow x = 1, - 1\]
\[\text { But } x < 0\]
\[ \Rightarrow x = - 1\]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{1}{x^3}\]
\[\text { At x } = - 1: \]
\[f''\left( - 1 \right) = \frac{2}{\left( - 1 \right)^3} = - 2 < 0\]
\[\text { So, x = - 1 is a point of local maximum }. \]
\[\text { Thus, the local maximum value is given by }\]
\[f\left( - 1 \right) = - 1 + \frac{1}{- 1} = - 1 - 1 = - 2\]
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