Find the point of local maxima and minima of $2 x^{3} - 3 x^{2} - 12 x + 12.$

x_{m a x} = 0, x_{m i n} = 1

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x_{m a x} = - 1, x_{m i n} = 1

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x_{m a x} = - 1, x_{m i n} = 2

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x_{m a x} = 2, x_{m i n} = - 2

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Solution

The correct option is D $x_{m a x} = - 1, x_{m i n} = 2$
Let $y = 2 x^{3} - 3 x^{2} - 12 x + 12$
$\frac{d y}{d x} = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2)$
For maximum or minimum ,
$\frac{d y}{d x} = 0$
$\Rightarrow x^{2} - x - 2 = 0$
$\Rightarrow (x - 2) (x + 1) = 0$
$\Rightarrow x = 2, - 1$
Now, $\frac{d^{2} y}{d x^{2}} = 12 x - 6$
At $x = 2$ , $\frac{d^{2} y}{d x^{2}} = 18 > 0$
Hence, $y$ is minimum at $x = 2$
At $x = - 1$ , $\frac{d^{2} y}{d x^{2}} = - 18 < 0$
Hence, $y$ is maximum at $x = - 1$

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