What is the extreme 2x^3-x^2-4x+8 ?

The extreme 2x^3-x^2-4x+8 is Maximum(-2/3 , 260/27),Minimum(1,5)

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extreme 2x^3-x^2-4x+8

extreme points

2 x^{3} - x^{2} - 4 x + 8

Maximum (- \frac{2}{3}, \frac{260}{27}), Minimum (1, 5)

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 + 8 : \frac{260}{27}

Maximum (- \frac{2}{3}, \frac{260}{27})

Plug

x = 1

into

2 x^{3} - x^{2} - 4 x + 8 : 5

Minimum (1, 5)

Maximum (- \frac{2}{3}, \frac{260}{27}), Minimum (1, 5)

f (x) = x^{3} - 13 x^{2} - 9 x

3 x^{3} - 9 x + 9 x y^{2}

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

f (x, y) = \frac{1}{x} + \frac{1}{y} + x y

f (x) = 7 (x - e^{x})

What is the extreme 2x^3-x^2-4x+8 ?
The extreme 2x^3-x^2-4x+8 is Maximum(-2/3 , 260/27),Minimum(1,5)