Question

f(x) = \[x^3 - 2a x^2 + a^2 x, a > 0, x \in R\] .

Answer 1

\[ f\left( x \right) = x^3 - 2a x^2 + a^2 x\]

\[ \Rightarrow f'\left( x \right) = 3 x^2 - 4ax + a^2 \]

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

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

\[ \Rightarrow 3 x^2 - 4ax + a^2 = 0\]

\[ \Rightarrow 3 x^2 - 3ax - ax + a^2 = 0\]

\[ \Rightarrow 3x\left( x - a \right) - a\left( x - a \right) = 0\]

\[ \Rightarrow \left( 3x - a \right)\left( x - a \right) = 0\]

\[ \Rightarrow x = a \text { and } \frac{a}{3}\]

\[\text {  Thus, x = a and x } = \frac{a}{3}\text {  are the possible points of local maxima or local minima }. \]

\[\text { Now,} \]

\[f''\left( x \right) = 6x - 4a\]

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

\[f''\left( a \right) = 6\left( a \right) - 4a = 2a > 0\]

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

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

\[f\left( a \right) = a^3 - 2a \left( a \right)^2 + a^2 \left( a \right) = 0\]

\[\text { At }x = \frac{a}{3}: \]

\[ f''\left( \frac{a}{3} \right) = 6\left( \frac{a}{3} \right) - 4a = - 2a < 0\]

\[\text { So }, x = \frac{a}{3} \text { is the point of local maximum }. \]

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

\[f\left( \frac{a}{3} \right) = \left( \frac{a}{3} \right)^3 - 2a \left( \frac{a}{3} \right)^2 + a^2 \left( \frac{a}{3} \right) = \frac{a^3}{27} - \frac{2 a^3}{9} + \frac{a^3}{3} = \frac{4 a^3}{27}\]