The minimum value of $2 x^{3} - 9 x^{2} + 12 x + 4$ is

4

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5

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6

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7

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8

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Solution

The correct option is E $8$
Let $y = 2 x^{3} - 9 x^{2} + 12 x + 4$

On differentiating w.r.t $x$ we get

$\frac{d y}{d x} = 6 x^{2} - 18 x + 12$

$f^{'} (x) = 0$

$∴ 6 x^{2} - 18 x + 12 = 0$

$\Rightarrow x^{2} - 3 x + 2 = 0$

$\Rightarrow x = 2 o r 1$

Now, $\frac{d^{2} y}{d x^{2}} = 12 x - 18 > 0$

or $x = 2$

$∴$ $x = 2$ is a point of local minima and $y_{m i n} = 16 - 36 + 24 + 4 = 8$

Suggest Corrections

Similar questions

2 x^{3} - 3 x^{2} - 12 x + 8

I = 2 \int 1 \frac{1}{\sqrt{2 x^{3} - 9 x^{2} + 12 x + 4}} d x

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 4

x =

f (x) = \frac{3}{10} x^{4} - \frac{4}{5} x^{3} - 3 x^{2} + \frac{36}{5} x + 11

f (x) = \frac{x^{4}}{4} + \frac{2}{3} x^{3} - \frac{5}{2} x^{2} - 6 x + 7

f (x) = {\{x (x - 2)\}}^{2}

2 x^{3} - 9 x^{2} + 12 x + 4

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