What is the extreme y=x^3-8x^2-12x+8 ?

The extreme y=x^3-8x^2-12x+8 is Maximum(-2/3 , 328/27),Minimum(6,-136)

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extreme y=x^3-8x^2-12x+8

extreme points

y = x^{3} - 8 x^{2} - 12 x + 8

Maximum (- \frac{2}{3}, \frac{328}{27}), Minimum (6, - 136)

Solution steps

f^{'} (x) = 3 x^{2} - 16 x - 12

Find intervals:

Increasing

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

Decreasing

: - \frac{2}{3} < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} - 8 x^{2} - 12 x + 8 : \frac{328}{27}

Maximum (- \frac{2}{3}, \frac{328}{27})

Plug

x = 6

into

x^{3} - 8 x^{2} - 12 x + 8 : - 136

Minimum (6, - 136)

Maximum (- \frac{2}{3}, \frac{328}{27}), Minimum (6, - 136)

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

f (x, y) = (2 x - x^{2}) (2 y - y^{2})

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

e^{7 x} + e^{- x}

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

What is the extreme y=x^3-8x^2-12x+8 ?
The extreme y=x^3-8x^2-12x+8 is Maximum(-2/3 , 328/27),Minimum(6,-136)