What is the extreme f(x)=(2x^2)/(x+1) ?

The extreme f(x)=(2x^2)/(x+1) is Maximum(-2,-8),Minimum(0,0)

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extreme f(x)=(2x^2)/(x+1)

extreme points

f (x) = \frac{2 x ^{2}}{x + 1}

Maximum (- 2, - 8), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{2 ( x ^{2} + 2 x )}{( x + 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

\frac{2 x ^{2}}{x + 1} : - 8

Maximum (- 2, - 8)

Plug

x = 0

into

\frac{2 x ^{2}}{x + 1} : 0

Minimum (0, 0)

Maximum (- 2, - 8), Minimum (0, 0)

f (x) = 5 - x^{2}

f (x) = x - x

f (x) = (\frac{ln ( x )}{x ^{2}})

f (x) = x^{3} - 4 x^{2} - 11 x + 30

f (x) = x^{2} - 8 x + 16

What is the extreme f(x)=(2x^2)/(x+1) ?
The extreme f(x)=(2x^2)/(x+1) is Maximum(-2,-8),Minimum(0,0)