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

The extreme f(x)=(x^2)/(x+2) is Maximum(-4,-8),Minimum(0,0)

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

extreme points

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

Maximum (- 4, - 8), Minimum (0, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 4

into

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

Maximum (- 4, - 8)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 4, - 8), Minimum (0, 0)

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

f (x) = x^{2} - 12 x - 13

f (x) = 15 x^{\frac{2}{3}} - 10 x

x^{4} + 4 x^{3} - 2

f (x) = 3 x^{2} ln (x)

What is the extreme f(x)=(x^2)/(x+2) ?
The extreme f(x)=(x^2)/(x+2) is Maximum(-4,-8),Minimum(0,0)