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, 28/3)

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

extreme points

f (x) = (\frac{1}{12} x^{3} + \frac{- 1}{2} x^{2} + 12)

Maximum (0, 12), Minimum (4, \frac{28}{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{28}{3}

Minimum (4, \frac{28}{3})

Maximum (0, 12), Minimum (4, \frac{28}{3})

y = x^{2} + 4 x - 7

f (x) = \frac{4}{3} x^{\frac{3}{2}} - \frac{1}{2} x^{2}

y = (x^{2} - 9)^{3}

f (x) = \frac{1}{x}, [1, \infty]

f (x) = - 0.1 x^{2} + 1.4 x + 98.9

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, 28/3)