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

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

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

extreme points

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

Maximum (- 2, - 1), Minimum (0, 0)

Solution steps

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

Maximum (- 2, - 1)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 2, - 1), Minimum (0, 0)

f (x) = x^{2} + 10

y = - 7 x + 21, (9, - 1)

f (x) = 6 x + 5

x^{3} - 64

f (x) = lo g_{9} (x)

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