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

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

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

extreme points

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

Minimum (- 2, 22 - 1), Maximum (2, - 1 - 22)

Solution steps

f^{'} (x) = - \frac{x ^{2} - 2}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x ^{3} + x - 2}{x - x ^{2}} : 22 - 1

Minimum (- 2, 22 - 1)

Plug

x = 2

into

\frac{x ^{3} + x - 2}{x - x ^{2}} : - 1 - 22

Maximum (2, - 1 - 22)

Minimum (- 2, 22 - 1), Maximum (2, - 1 - 22)

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