What is the extreme f(x)=2x^3-3x^2-12x+18 ?

The extreme f(x)=2x^3-3x^2-12x+18 is Maximum(-1,25),Minimum(2,-2)

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extreme f(x)=2x^3-3x^2-12x+18

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 18

Maximum (- 1, 25), Minimum (2, - 2)

Solution steps

f^{'} (x) = 6 x^{2} - 6 x - 12

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

2 x^{3} - 3 x^{2} - 12 x + 18 : 25

Maximum (- 1, 25)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 18 : - 2

Minimum (2, - 2)

Maximum (- 1, 25), Minimum (2, - 2)

f (x) = x^{3} - 3 x^{2} - 9 x + 1, 0 \leq x \leq 5

f (x, y) = \frac{1}{4} x^{4} - \frac{1}{2} x^{2} + y^{2}

f (x) = x + \frac{4}{x ^{2}}

f (x) = (x - 2) (x + 1) (x + 4)

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

What is the extreme f(x)=2x^3-3x^2-12x+18 ?
The extreme f(x)=2x^3-3x^2-12x+18 is Maximum(-1,25),Minimum(2,-2)