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Question
f(x) = (x - 1) (x + 2)2.
Solution
\[\text { Given }: \hspace{0.167em} f\left( x \right) = \left( x - 1 \right) \left( x + 2 \right)^2 \]
\[ = \left( x - 1 \right)\left( x^2 + 4x + 4 \right)\]
\[ = x^3 + 4 x^2 + 4x - x^2 - 4x - 4\]
\[ = x^3 + 3 x^2 - 4\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 + 6x\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 3 x^2 + 6x = 0\]
\[ \Rightarrow 3x\left( x + 2 \right) = 0\]
\[ \Rightarrow x = 0 \text { and } - 2\]
\[\text { Thus, x = 0 and x = - 2 are the possible points of local maxima or local minima }. \]
\[\text { Now }, \]
\[f''\left( x \right) = 6x + 6\]
\[\text { At x } = 0: \]
\[ f''\left( 0 \right) = 6\left( 0 \right) + 6 = 6 > 0\]
\[\text { So, x = 0 is the point of local minimum } . \]
\[\text { The local minimum value is given by }\]
\[f\left( 0 \right) = \left( 0 - 1 \right) \left( 0 + 2 \right)^2 = - 4\]
\[\text { At }x = - 2: \]
\[ f''\left( - 2 \right) = 6\left( - 2 \right) + 6 = - 6 < 0\]
\[\text { So, x = - 2 is the point of local maximum} . \]
\[\text { The local maximum value is given by } \]
\[f\left( - 2 \right) = \left( - 2 - 1 \right) \left( - 2 + 2 \right)^2 = 0\]
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