The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$ has

No maxima and minima

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One maxima

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One minima

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Two minima

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Solution

The correct options are
B One maxima
C One minima
Given function is
$f (x) = 2 x^{3} - 3 x - 12 x + 4$

$f^{1} (x) = 6 x^{2} - 6 x - 12$

$f^{1} (x) = 6 (x^{2} - x - 2)$

$= 6 (x - 2) (x + 1)$

Capital points $x = - 1$ and $x = 2$

$f^{11} (x) = 6 (2 x - 1)$

$f^{11} (2) = 6 (4 - 1) = 18 > 0$
Local minimum

$f^{11} (- 1) = 6 (- 2 - 1) = - 18 < 0$
Local maximum

There are only one local maxima and minima.

Suggest Corrections

Similar questions

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

f (x) = 2 x^{3} - 3 x^{2} - 36 x + 2

f (x) = 2 x^{3} - 15 x^{2} + 36 x + 10

f (x) = 2 x^{3} - 3 x^{2_{-}} 12 x + 8

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 8, f^{'} (2) = 0

f^{''} (2) > 0

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

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