What is the extreme f(x)=2x^3-x^2-4x+12 ?

The extreme f(x)=2x^3-x^2-4x+12 is Maximum(-2/3 , 368/27),Minimum(1,9)

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extreme f(x)=2x^3-x^2-4x+12

extreme points

f (x) = 2 x^{3} - x^{2} - 4 x + 12

Maximum (- \frac{2}{3}, \frac{368}{27}), Minimum (1, 9)

Solution steps

f^{'} (x) = 6 x^{2} - 2 x - 4

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - \frac{2}{3}

into

2 x^{3} - x^{2} - 4 x + 12 : \frac{368}{27}

Maximum (- \frac{2}{3}, \frac{368}{27})

Plug

x = 1

into

2 x^{3} - x^{2} - 4 x + 12 : 9

Minimum (1, 9)

Maximum (- \frac{2}{3}, \frac{368}{27}), Minimum (1, 9)

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

f (x) = x^{3} - 3 x^{2} + 7 x - 10

f (x) = - 10 x^{2} - 40 x - 2

f (x) = 4 x^{2} + y^{2} - 8

10 x^{2} - \frac{250}{x}

What is the extreme f(x)=2x^3-x^2-4x+12 ?
The extreme f(x)=2x^3-x^2-4x+12 is Maximum(-2/3 , 368/27),Minimum(1,9)