What is the extreme f(x)=2x^3-3x^2-12x+15 ?

The extreme f(x)=2x^3-3x^2-12x+15 is Maximum(-1,22),Minimum(2,-5)

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extreme f(x)=2x^3-3x^2-12x+15

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 15

Maximum (- 1, 22), Minimum (2, - 5)

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 + 15 : 22

Maximum (- 1, 22)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 15 : - 5

Minimum (2, - 5)

Maximum (- 1, 22), Minimum (2, - 5)

f (x, y) = 2 x - 3 y + 6

f (x, y) = ln (4 - x - y)

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

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

f (x) = x 9 - x^{2}

What is the extreme f(x)=2x^3-3x^2-12x+15 ?
The extreme f(x)=2x^3-3x^2-12x+15 is Maximum(-1,22),Minimum(2,-5)