What is the extreme f(x)=(x^2(x-1))/(x+2),x\ne 2 ?

The extreme f(x)=(x^2(x-1))/(x+2),x\ne 2 is

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extreme f(x)=(x^2(x-1))/(x+2),x\ne 2

extreme

Minimum (\frac{- 5 - 57}{4}, \frac{19 57 + 143}{8}), Maximum (0, 0), Minimum (\frac{- 5 + 57}{4}, - \frac{19 57 - 143}{8})

Solution steps

Find the critical points:

x = \frac{- 5 - 57}{4}, x = 0, x = \frac{- 5 + 57}{4}

Domain of

\frac{x ^{2} ( x - 1 )}{x + 2}, x \neq = 2 : x < - 2 or - 2 < x < 2 or x > 2

Find intervals:

Decreasing

: - \infty < x < \frac{- 5 - 57}{4},

Increasing

: \frac{- 5 - 57}{4} < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \frac{- 5 + 57}{4},

Increasing

: \frac{- 5 + 57}{4} < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = \frac{- 5 - 57}{4}

into

\frac{x ^{2} ( x - 1 )}{x + 2} \Rightarrow y = \frac{19 57 + 143}{8}

Minimum (\frac{- 5 - 57}{4}, \frac{19 57 + 143}{8})

Plug the extreme point

x = 0

into

\frac{x ^{2} ( x - 1 )}{x + 2} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = \frac{- 5 + 57}{4}

into

\frac{x ^{2} ( x - 1 )}{x + 2} \Rightarrow y = - \frac{19 57 - 143}{8}

Minimum (\frac{- 5 + 57}{4}, - \frac{19 57 - 143}{8})

Minimum (\frac{- 5 - 57}{4}, \frac{19 57 + 143}{8}), Maximum (0, 0), Minimum (\frac{- 5 + 57}{4}, - \frac{19 57 - 143}{8})

What is the extreme f(x)=(x^2(x-1))/(x+2),x\ne 2 ?
The extreme f(x)=(x^2(x-1))/(x+2),x\ne 2 is