What is the extreme f(x)=(x^2)/(x+3) ?

The extreme f(x)=(x^2)/(x+3) is Maximum(-6,-12),Minimum(0,0)

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extreme f(x)=(x^2)/(x+3)

extreme points

f (x) = \frac{x ^{2}}{x + 3}

Maximum (- 6, - 12), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{x ^{2} + 6 x}{( x + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 6

into

\frac{x ^{2}}{x + 3} : - 12

Maximum (- 6, - 12)

Plug

x = 0

into

\frac{x ^{2}}{x + 3} : 0

Minimum (0, 0)

Maximum (- 6, - 12), Minimum (0, 0)

x

5

K (r, s) = 8 r - s

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 6 x

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

What is the extreme f(x)=(x^2)/(x+3) ?
The extreme f(x)=(x^2)/(x+3) is Maximum(-6,-12),Minimum(0,0)