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Question
f(x) = \[x^3 - 2a x^2 + a^2 x, a > 0, x \in R\] .
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Solution
\[ f\left( x \right) = x^3 - 2a x^2 + a^2 x\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 - 4ax + a^2 \]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 3 x^2 - 4ax + a^2 = 0\]
\[ \Rightarrow 3 x^2 - 3ax - ax + a^2 = 0\]
\[ \Rightarrow 3x\left( x - a \right) - a\left( x - a \right) = 0\]
\[ \Rightarrow \left( 3x - a \right)\left( x - a \right) = 0\]
\[ \Rightarrow x = a \text { and } \frac{a}{3}\]
\[\text { Thus, x = a and x } = \frac{a}{3}\text { are the possible points of local maxima or local minima }. \]
\[\text { Now,} \]
\[f''\left( x \right) = 6x - 4a\]
\[\text { At } x = a: \]
\[f''\left( a \right) = 6\left( a \right) - 4a = 2a > 0\]
\[\text { So, x = a is the point of local minimum }. \]
\[\text { The local minimum value is given by }\]
\[f\left( a \right) = a^3 - 2a \left( a \right)^2 + a^2 \left( a \right) = 0\]
\[\text { At }x = \frac{a}{3}: \]
\[ f''\left( \frac{a}{3} \right) = 6\left( \frac{a}{3} \right) - 4a = - 2a < 0\]
\[\text { So }, x = \frac{a}{3} \text { is the point of local maximum }. \]
\[\text { The local maximum value is given by }\]
\[f\left( \frac{a}{3} \right) = \left( \frac{a}{3} \right)^3 - 2a \left( \frac{a}{3} \right)^2 + a^2 \left( \frac{a}{3} \right) = \frac{a^3}{27} - \frac{2 a^3}{9} + \frac{a^3}{3} = \frac{4 a^3}{27}\]
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