Question

f(x) =  (x \[-\] 1) (x+2)². 

Answer 1

\[\text { Given }: f\left( x \right) = \left( x - 1 \right) \left( x + 2 \right)^2 \]

\[ \Rightarrow f'\left( x \right) = \left( x + 2 \right)^2 + 2\left( x + 2 \right)\left( x - 1 \right)\]

\[\text{ For a local maximum or a local minimum, we must have }\]

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

\[ \Rightarrow \left( x + 2 \right)^2 + 2\left( x + 2 \right)\left( x - 1 \right) = 0\]

\[ \Rightarrow \left( x + 2 \right)\left( x + 2 + 2x - 2 \right) = 0\]

\[ \Rightarrow \left( x + 2 \right)\left( 3x \right) = 0\]

\[ \Rightarrow x = 0, - 2\]

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

\[\left( - 2 - 1 \right) \left( - 2 + 2 \right)^2 = 0\]