What is the extreme f(x)=(1/12 x^3-1/2 x^2-12) ?

The extreme f(x)=(1/12 x^3-1/2 x^2-12) is Maximum(0,-12),Minimum(4,-44/3)

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extreme f(x)=(1/12 x^3-1/2 x^2-12)

extreme

Maximum (0, - 12), Minimum (4, - \frac{44}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

\frac{1}{12} x^{3} - \frac{1}{2} x^{2} - 12 : - 12

Maximum (0, - 12)

Plug

x = 4

into

\frac{1}{12} x^{3} - \frac{1}{2} x^{2} - 12 : - \frac{44}{3}

Minimum (4, - \frac{44}{3})

Maximum (0, - 12), Minimum (4, - \frac{44}{3})

What is the extreme f(x)=(1/12 x^3-1/2 x^2-12) ?
The extreme f(x)=(1/12 x^3-1/2 x^2-12) is Maximum(0,-12),Minimum(4,-44/3)