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

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

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

extreme points

y = \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)

\frac{( x - 1 ) ( x + 4 )}{( x + 1 ) ( x - 6 )}

f (x) = 1

f (x) = 8 x^{4}

\frac{1}{x ^{2} - 6 x + 14}

y = 2 x^{2} - x + 2

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