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

The extreme f(x)=((x^2)/(x+1)) is Maximum(-2,-4),Minimum(0,0)

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

extreme points

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

Maximum (- 2, - 4), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{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{x ^{2}}{x + 1} : - 4

Maximum (- 2, - 4)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 2, - 4), Minimum (0, 0)

P (x, y) = - 3 x + 4 y

\frac{3 ^{x} + 5}{9}

\frac{x}{x - 2 x}

f (x) = x^{3} - 3 x + 1, 0 \leq x \leq 2

f (x, y) = y^{3} + x^{3} - \frac{21}{2} y^{2} - 3 x + 30 y

What is the extreme f(x)=((x^2)/(x+1)) ?
The extreme f(x)=((x^2)/(x+1)) is Maximum(-2,-4),Minimum(0,0)