MCQ
Local maximum and local minimum values of the function $(x - 1){(x + 2)^2}$ are
- A$-4, 0$
- ✓$0, -4$
- C$4, 0$
- DNone of these
$f'(x) = 0$ ==>$x = 0,\, - 2$
$f( - 2) = ( - 2 - 1){( - 2 + 2)^2} = 0$ (Maximum value)
and $f(0) = (0 - 1){(0 + 2)^2} = - 4$ (Minimum value).
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| Column | Maximum of $z$ |
| $A$ | $300$ |
| $B$ | $325$ |