What is the extreme f(x)=2x^3-3x^2-12x+1 ?

The extreme f(x)=2x^3-3x^2-12x+1 is Maximum(-1,8),Minimum(2,-19)

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extreme f(x)=2x^3-3x^2-12x+1

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 1

Maximum (- 1, 8), Minimum (2, - 19)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

2 x^{3} - 3 x^{2} - 12 x + 1 : 8

Maximum (- 1, 8)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 1 : - 19

Minimum (2, - 19)

Maximum (- 1, 8), Minimum (2, - 19)

f (x, y) = x y

f (x) = x^{4} + x^{3} - 6 x^{2}

f (x) = 9 - x^{2} - y^{2}

f (x, y) = x + y

f (x, y) = 16 - x^{2} - y^{2}

What is the extreme f(x)=2x^3-3x^2-12x+1 ?
The extreme f(x)=2x^3-3x^2-12x+1 is Maximum(-1,8),Minimum(2,-19)