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

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

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

extreme points

f (x) = 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)

y = x + 3

(3, 6), (7, 7)

f (x) = 3 cos (x) - 2

g (x) = x^{2} ∣ x ∣ + 5

f (x) = \frac{2 x + 1}{x - 5}

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