What is the extreme f(x)=(x^2)/(2x+4) ?

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

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extreme f(x)=(x^2)/(2x+4)

extreme points

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

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

Solution steps

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

Maximum (- 4, - 4)

Plug

x = 0

into

\frac{x ^{2}}{2 x + 4} : 0

Minimum (0, 0)

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

f (x, y) = y x e^{- (x^{2} + y^{2})}

f (x, y) = - 2 x^{4} + y^{2} + x^{2} - 2 y

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

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

F (I, J) = 25 I N + 12 J N

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