#### Question

f(x) = x^{3}\[-\] 6x^{2} + 9x + 15

#### Solution

\[\text { Given }: f\left( x \right) = x^3 - 6 x^2 + 9x + 15\]

\[ \Rightarrow f'\left( x \right) = 3 x^2 - 12x + 9\]

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

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

\[ \Rightarrow 3 x^2 - 12x + 9 = 0\]

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

\[ \Rightarrow \left( x - 1 \right)\left( x - 3 \right) = 0\]

\[ \Rightarrow x = 1 \text { and } 3\]

\[\text { Thus, x = 1 and x = 3 are the possible points of local maxima or local minima } . \]

\[\text { Now,} \]

\[f''\left( x \right) = 6x - 12\]

\[\text { At }x = 1: \]

\[ f''\left( 1 \right) = 6\left( 1 \right) - 12 = - 6 < 0\]

\[\text {So, x = 1 is the point of local maximum } . \]

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

\[f\left( 1 \right) = 1^3 - 6 \left( 1 \right)^2 + 9 \times 1 + 15 = 19\]

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

\[ f''\left( 3 \right) = 6\left( 3 \right) - 12 = 6 > 0\]

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

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

\[f\left( 3 \right) = 3^3 - 6 \left( 3 \right)^2 + 9 \times 3 + 15 = 15\]