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Question
f(x) = (x \[-\] 1) (x+2)2.
Solution
\[\text { Given }: f\left( x \right) = \left( x - 1 \right) \left( x + 2 \right)^2 \]
\[ \Rightarrow f'\left( x \right) = \left( x + 2 \right)^2 + 2\left( x + 2 \right)\left( x - 1 \right)\]
\[\text{ For a local maximum or a local minimum, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \left( x + 2 \right)^2 + 2\left( x + 2 \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow \left( x + 2 \right)\left( x + 2 + 2x - 2 \right) = 0\]
\[ \Rightarrow \left( x + 2 \right)\left( 3x \right) = 0\]
\[ \Rightarrow x = 0, - 2\]
Since f '(x) changes from negative to positive when x increases through 0, x = 0 is the point of local minima.
The local minimum value of f (x) at x = 0 is given by \[\left( 0 - 1 \right) \left( 0 + 2 \right)^2 = - 4\]
Since f '(x) changes sign from positive to negative when x increases through \[- 2\] ,x = \[- 2\] is the point of local maxima.
The local maximum value of f (x) at x = \[- 2\] is given by
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