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

The extreme f(x)=(x^2)/(x+4) is Maximum(-8,-16),Minimum(0,0)

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

extreme points

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

Maximum (- 8, - 16), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{x ^{2} + 8 x}{( x + 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 8

into

\frac{x ^{2}}{x + 4} : - 16

Maximum (- 8, - 16)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 8, - 16), Minimum (0, 0)

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

\frac{ln ( x ^{2} )}{x}

t - 3 t

f (x) = 7 x x - x^{2}

f (x) = 3 x^{4} - 12 x^{3}

What is the extreme f(x)=(x^2)/(x+4) ?
The extreme f(x)=(x^2)/(x+4) is Maximum(-8,-16),Minimum(0,0)