#### Question

f(x) = \[x + \frac{a2}{x}, a > 0,\] , x ≠ 0 .

#### Solution

\[\text { Given }: f\left( x \right) = x + \frac{a^2}{x}\]

\[ \Rightarrow f'\left( x \right) = 1 - \frac{a^2}{x^2}\]

\[\text { For the local maxima or minima, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 1 - \frac{a^2}{x^2} = 0\]

\[ \Rightarrow x^2 = a^2 \]

\[ \Rightarrow x = \pm a \]

\[\text { Thus, x = a and x = - a are the possible points of local maxima or local minima }. \]

\[\text { Now,} \]

\[ f''\left( x \right) = \frac{a^2}{x^3}\]

\[\text { At x = a }: \]

\[ f''\left( a \right) = \frac{a^2}{\left( a \right)^3} = \frac{1}{a} > 0\]

\[\text { So, x = a is the point of local minimum } . \]

\[\text { The local minimum value is given by }\]

\[f\left( a \right) = x + \frac{a^2}{x} = a + a = 2a\]

\[At x = - a: \]

\[ f''\left( a \right) = \frac{a^2}{\left( - a \right)^3} = - \frac{1}{a} < 0\]

\[\text { So, x = - a is the point of local maximum }. \]

\[\text { The local maximum value is given by }\]

\[f\left( - a \right) = x + \frac{a^2}{x} = - a - a = - 2a\]