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

The extreme f(x)=2x^3-3x^2-36x is Maximum(-2,44),Minimum(3,-81)

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

extreme points

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

Maximum (- 2, 44), Minimum (3, - 81)

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 : 44

Maximum (- 2, 44)

Plug

x = 3

into

2 x^{3} - 3 x^{2} - 36 x : - 81

Minimum (3, - 81)

Maximum (- 2, 44), Minimum (3, - 81)

(1.3) (5.1)

f (x) = x^{3} - 6 x^{2} + 15

f (x) = 3 sin (4 x)

h (x) = \frac{2}{( 2 x + 1 )}

f (x) = 2 x + 1

What is the extreme f(x)=2x^3-3x^2-36x ?
The extreme f(x)=2x^3-3x^2-36x is Maximum(-2,44),Minimum(3,-81)