What is the extreme f(x)=2x^3-x^2-4x+8(-1) ?

The extreme f(x)=2x^3-x^2-4x+8(-1) is Maximum(-2/3 ,-172/27),Minimum(1,-11)

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extreme f(x)=2x^3-x^2-4x+8(-1)

extreme points

f (x) = 2 x^{3} - x^{2} - 4 x + 8 (- 1)

Maximum (- \frac{2}{3}, - \frac{172}{27}), Minimum (1, - 11)

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 (- 1) : - \frac{172}{27}

Maximum (- \frac{2}{3}, - \frac{172}{27})

Plug

x = 1

into

2 x^{3} - x^{2} - 4 x + 8 (- 1) : - 11

Minimum (1, - 11)

Maximum (- \frac{2}{3}, - \frac{172}{27}), Minimum (1, - 11)

f (x, y) = 2 x^{5} + x^{3} y^{2} + 6 x y^{4}

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

f (x, y) = x^{2} y^{3} - (x - 2 y)^{2} + 6 x^{4} y = 3 x y

f (x) = 2 x^{4} - 20 x^{2} + 18

- x^{2} + 4 x + 6

What is the extreme f(x)=2x^3-x^2-4x+8(-1) ?
The extreme f(x)=2x^3-x^2-4x+8(-1) is Maximum(-2/3 ,-172/27),Minimum(1,-11)