The maximum value of $f (x) = (x - 2)^{2} (x - 3)$ is

2

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4

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- 4

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0

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Solution

The correct option is B $0$
$f (x) = (x - 2)^{2} (x - 3)$
$f^{'} (x) = (x - 2) (3 x - 8)$
For maxima or minima,
$f^{'} (x) = 0$
$\Rightarrow x = 2, \frac{8}{3}$
$f^{''} (x) = 6 x - 14$
$f^{''} (2) < 0$
$\Rightarrow$ f(x) has a maximum at x=2.
$f (2) = 0$

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