What is the extreme f(x)=2x^3+3x^2-36x+4 ?

The extreme f(x)=2x^3+3x^2-36x+4 is Maximum(-3,85),Minimum(2,-40)

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extreme f(x)=2x^3+3x^2-36x+4

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 4

Maximum (- 3, 85), Minimum (2, - 40)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 3

into

2 x^{3} + 3 x^{2} - 36 x + 4 : 85

Maximum (- 3, 85)

Plug

x = 2

into

2 x^{3} + 3 x^{2} - 36 x + 4 : - 40

Minimum (2, - 40)

Maximum (- 3, 85), Minimum (2, - 40)

x^{3} + 2 x^{2} + x + 2

f (x) = x^{3} - 18 x^{2} + 81 x

y = \frac{x ^{2}}{x ^{2} - 1}

f (x) = x^{2} - 4 x y + 5

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

What is the extreme f(x)=2x^3+3x^2-36x+4 ?
The extreme f(x)=2x^3+3x^2-36x+4 is Maximum(-3,85),Minimum(2,-40)