What is the extreme f(x)=(4x^2)/(3+x) ?

The extreme f(x)=(4x^2)/(3+x) is Maximum(-6,-48),Minimum(0,0)

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extreme f(x)=(4x^2)/(3+x)

extreme

Maximum (- 6, - 48), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{4 ( 6 x + x ^{2} )}{( 3 + x ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 6

into

\frac{4 x ^{2}}{3 + x} : - 48

Maximum (- 6, - 48)

Plug

x = 0

into

\frac{4 x ^{2}}{3 + x} : 0

Minimum (0, 0)

Maximum (- 6, - 48), Minimum (0, 0)

What is the extreme f(x)=(4x^2)/(3+x) ?
The extreme f(x)=(4x^2)/(3+x) is Maximum(-6,-48),Minimum(0,0)