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Question
f(x) = x3\[-\] 6x2 + 9x + 15
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Solution
\[\text { Given }: f\left( x \right) = x^3 - 6 x^2 + 9x + 15\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 - 12x + 9\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 3 x^2 - 12x + 9 = 0\]
\[ \Rightarrow x^2 - 4x + 3 = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x - 3 \right) = 0\]
\[ \Rightarrow x = 1 \text { and } 3\]
\[\text { Thus, x = 1 and x = 3 are the possible points of local maxima or local minima } . \]
\[\text { Now,} \]
\[f''\left( x \right) = 6x - 12\]
\[\text { At }x = 1: \]
\[ f''\left( 1 \right) = 6\left( 1 \right) - 12 = - 6 < 0\]
\[\text {So, x = 1 is the point of local maximum } . \]
\[\text { The local maximum value is given by }\]
\[f\left( 1 \right) = 1^3 - 6 \left( 1 \right)^2 + 9 \times 1 + 15 = 19\]
\[\text { At }x = 3: \]
\[ f''\left( 3 \right) = 6\left( 3 \right) - 12 = 6 > 0\]
\[\text { So, x = 3 is the point of local minimum }. \]
\[\text { The local minimum value is given by }\]
\[f\left( 3 \right) = 3^3 - 6 \left( 3 \right)^2 + 9 \times 3 + 15 = 15\]
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