What is the extreme f(x)=2x^3-3x^2-36x+5 ?

The extreme f(x)=2x^3-3x^2-36x+5 is Maximum(-2,49),Minimum(3,-76)

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extreme f(x)=2x^3-3x^2-36x+5

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 36 x + 5

Maximum (- 2, 49), Minimum (3, - 76)

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 + 5 : 49

Maximum (- 2, 49)

Plug

x = 3

into

2 x^{3} - 3 x^{2} - 36 x + 5 : - 76

Minimum (3, - 76)

Maximum (- 2, 49), Minimum (3, - 76)

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

- 2 x + 5 y = 10

(2, 3), (- 3, 15)

2 x + 3 y = 8

f (x) = 4

What is the extreme f(x)=2x^3-3x^2-36x+5 ?
The extreme f(x)=2x^3-3x^2-36x+5 is Maximum(-2,49),Minimum(3,-76)