What is the extreme y=2x^3+3x^2-12x ?

The extreme y=2x^3+3x^2-12x is Maximum(-2,20),Minimum(1,-7)

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extreme y=2x^3+3x^2-12x

extreme points

y = 2 x^{3} + 3 x^{2} - 12 x

Maximum (- 2, 20), Minimum (1, - 7)

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 : 20

Maximum (- 2, 20)

Plug

x = 1

into

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

Minimum (1, - 7)

Maximum (- 2, 20), Minimum (1, - 7)

f (x) = y - x

f (x, y) = e^{x - y}

f (x, y) = \frac{1}{2} x^{4} - 3 x y + 3 y^{4}

3 x^{2}

f (x) = x^{2} - 12 x + 10

What is the extreme y=2x^3+3x^2-12x ?
The extreme y=2x^3+3x^2-12x is Maximum(-2,20),Minimum(1,-7)