What is the extreme f(x)=2x^3-x^2-4x+10 ?

The extreme f(x)=2x^3-x^2-4x+10 is Maximum(-2/3 , 314/27),Minimum(1,7)

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extreme f(x)=2x^3-x^2-4x+10

extreme points

f (x) = 2 x^{3} - x^{2} - 4 x + 10

Maximum (- \frac{2}{3}, \frac{314}{27}), Minimum (1, 7)

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 + 10 : \frac{314}{27}

Maximum (- \frac{2}{3}, \frac{314}{27})

Plug

x = 1

into

2 x^{3} - x^{2} - 4 x + 10 : 7

Minimum (1, 7)

Maximum (- \frac{2}{3}, \frac{314}{27}), Minimum (1, 7)

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

43 x^{3} - 48 x

f (x) = \frac{( 2 + x )}{x}

f (x) = \frac{x + 3}{x}

x^{3} + 5

What is the extreme f(x)=2x^3-x^2-4x+10 ?
The extreme f(x)=2x^3-x^2-4x+10 is Maximum(-2/3 , 314/27),Minimum(1,7)