What is the extreme f(x)=x^3-2x^2-4x+5 ?

The extreme f(x)=x^3-2x^2-4x+5 is Maximum(-2/3 , 175/27),Minimum(2,-3)

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extreme f(x)=x^3-2x^2-4x+5

extreme points

f (x) = x^{3} - 2 x^{2} - 4 x + 5

Maximum (- \frac{2}{3}, \frac{175}{27}), Minimum (2, - 3)

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 2 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} - 2 x^{2} - 4 x + 5 : \frac{175}{27}

Maximum (- \frac{2}{3}, \frac{175}{27})

Plug

x = 2

into

x^{3} - 2 x^{2} - 4 x + 5 : - 3

Minimum (2, - 3)

Maximum (- \frac{2}{3}, \frac{175}{27}), Minimum (2, - 3)

What is the extreme f(x)=x^3-2x^2-4x+5 ?
The extreme f(x)=x^3-2x^2-4x+5 is Maximum(-2/3 , 175/27),Minimum(2,-3)