What is the extreme 2x^3-x^2-4x+12 ?

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

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extreme 2x^3-x^2-4x+12

extreme points

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) = x^{3} - \frac{3}{2} x^{2} - 6 x + 1

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

f (x) = 600 + 35 x

f (x) = 3 cos (π x), 0 \leq x \leq \frac{1}{6}

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

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