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)

f (x) = 2 x^{4} - 3 x^{2} + 1

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

x^{3} + 3 x + 8

f (x) = \frac{1}{x + 11}

sec (x)

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)