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

The extreme f(x)=2x^3+3x^2-36x+1 is Maximum(-3,82),Minimum(2,-43)

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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 (- 3, 82), Minimum (2, - 43)

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 + 1 : 82

Maximum (- 3, 82)

Plug

x = 2

into

2 x^{3} + 3 x^{2} - 36 x + 1 : - 43

Minimum (2, - 43)

Maximum (- 3, 82), Minimum (2, - 43)

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

f (x) = x + e^{- 4 x}

f (x) = - 2 x^{3} + 9 x^{2} - 12 x

- \frac{5 x}{x ^{2} + 4}

f (x) = x^{2} + 6 x

What is the extreme f(x)=2x^3+3x^2-36x+1 ?
The extreme f(x)=2x^3+3x^2-36x+1 is Maximum(-3,82),Minimum(2,-43)