The maximum value of $f (x) = 2 x^{3} - 9 x^{2} + 12 x - 3$ in the interval $0 \leq x \leq 3$ is
6

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Solution

The correct option is A 6
$f (x) = 2 x^{3} - 9 x^{2} + 12 x - 3, x ϵ [0, 3]$
$f^{'} (x) = 6 x^{2} - 18 x + 12 = 0$
$x = 1, 2 ϵ (0, 3)$

$x$ $0$ $1$ $2$ $3$

$f (x)$ $- 3$ $2$ $1$ $6$

Maximum value of $f (x) = 6.$

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f (x) = 6 - 12 x + 9 x^{2} - 2 x^{3}, 1 \leq x \leq 4.

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