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

The extreme f(x)=2x^3-x^2-4x+2 is Maximum(-2/3 , 98/27),Minimum(1,-1)

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

extreme points

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

Maximum (- \frac{2}{3}, \frac{98}{27}), Minimum (1, - 1)

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 + 2 : \frac{98}{27}

Maximum (- \frac{2}{3}, \frac{98}{27})

Plug

x = 1

into

2 x^{3} - x^{2} - 4 x + 2 : - 1

Minimum (1, - 1)

Maximum (- \frac{2}{3}, \frac{98}{27}), Minimum (1, - 1)

f (x) = 2 x^{3} - 12 x^{2} + 18 x + 5

f (x) = 2 x \cdot ln (x)

f (x) = 5 x^{\frac{1}{3}} - x^{\frac{5}{3}}

6 x^{2} + 30 x - 84

y = r - (x - r)^{2}

What is the extreme f(x)=2x^3-x^2-4x+2 ?
The extreme f(x)=2x^3-x^2-4x+2 is Maximum(-2/3 , 98/27),Minimum(1,-1)