What is the extreme f(x)=2x^3-3x^2-36x+1 ?

The extreme f(x)=2x^3-3x^2-36x+1 is Maximum(-2,45),Minimum(3,-80)

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extreme f(x)=2x^3-3x^2-36x+1

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 36 x + 1

Maximum (- 2, 45), Minimum (3, - 80)

Solution steps

f^{'} (x) = 6 x^{2} - 6 x - 36

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 2

into

2 x^{3} - 3 x^{2} - 36 x + 1 : 45

Maximum (- 2, 45)

Plug

x = 3

into

2 x^{3} - 3 x^{2} - 36 x + 1 : - 80

Minimum (3, - 80)

Maximum (- 2, 45), Minimum (3, - 80)

f (x, y) = 2 x^{2} + 4 x y^{3} - 2 y^{2} + 15

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

f (x, y) = x^{2} + y^{2} + x y - 2 x - y + 1

f (x) = x x^{2} + 16

f (x) = \frac{x}{x ^{2} + 14 x + 48}

What is the extreme f(x)=2x^3-3x^2-36x+1 ?
The extreme f(x)=2x^3-3x^2-36x+1 is Maximum(-2,45),Minimum(3,-80)