What is the extreme f(x)=2x^3+3x^2-12x+7 ?

The extreme f(x)=2x^3+3x^2-12x+7 is Maximum(-2,27),Minimum(1,0)

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extreme f(x)=2x^3+3x^2-12x+7

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 7

Maximum (- 2, 27), Minimum (1, 0)

Solution steps

f^{'} (x) = 6 x^{2} + 6 x - 12

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

2 x^{3} + 3 x^{2} - 12 x + 7 : 27

Maximum (- 2, 27)

Plug

x = 1

into

2 x^{3} + 3 x^{2} - 12 x + 7 : 0

Minimum (1, 0)

Maximum (- 2, 27), Minimum (1, 0)

- x + 2 cos (x)

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

f (x) = cos (x) - 3 x

f (x) = 3 x^{2} - 24 ln (x)

f (x) = 5 sin (3 x) - 5, (0, 2 π)

What is the extreme f(x)=2x^3+3x^2-12x+7 ?
The extreme f(x)=2x^3+3x^2-12x+7 is Maximum(-2,27),Minimum(1,0)