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Question
f(x) = \[x + \frac{a2}{x}, a > 0,\] , x ≠ 0 .
Solution
\[\text { Given }: f\left( x \right) = x + \frac{a^2}{x}\]
\[ \Rightarrow f'\left( x \right) = 1 - \frac{a^2}{x^2}\]
\[\text { For the local maxima or minima, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 1 - \frac{a^2}{x^2} = 0\]
\[ \Rightarrow x^2 = a^2 \]
\[ \Rightarrow x = \pm a \]
\[\text { Thus, x = a and x = - a are the possible points of local maxima or local minima }. \]
\[\text { Now,} \]
\[ f''\left( x \right) = \frac{a^2}{x^3}\]
\[\text { At x = a }: \]
\[ f''\left( a \right) = \frac{a^2}{\left( a \right)^3} = \frac{1}{a} > 0\]
\[\text { So, x = a is the point of local minimum } . \]
\[\text { The local minimum value is given by }\]
\[f\left( a \right) = x + \frac{a^2}{x} = a + a = 2a\]
\[At x = - a: \]
\[ f''\left( a \right) = \frac{a^2}{\left( - a \right)^3} = - \frac{1}{a} < 0\]
\[\text { So, x = - a is the point of local maximum }. \]
\[\text { The local maximum value is given by }\]
\[f\left( - a \right) = x + \frac{a^2}{x} = - a - a = - 2a\]
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