What is the extreme f(x)=x^3-(1/2)x^2-4x+12 ?

The extreme f(x)=x^3-(1/2)x^2-4x+12 is Maximum(-1, 29/2),Minimum(4/3 , 220/27)

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extreme f(x)=x^3-(1/2)x^2-4x+12

extreme

Maximum (- 1, \frac{29}{2}), Minimum (\frac{4}{3}, \frac{220}{27})

Solution steps

f^{'} (x) = 3 x^{2} - x - 4

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = - 1

into

x^{3} - \frac{1}{2} x^{2} - 4 x + 12 : \frac{29}{2}

Maximum (- 1, \frac{29}{2})

Plug

x = \frac{4}{3}

into

x^{3} - \frac{1}{2} x^{2} - 4 x + 12 : \frac{220}{27}

Minimum (\frac{4}{3}, \frac{220}{27})

Maximum (- 1, \frac{29}{2}), Minimum (\frac{4}{3}, \frac{220}{27})

What is the extreme f(x)=x^3-(1/2)x^2-4x+12 ?
The extreme f(x)=x^3-(1/2)x^2-4x+12 is Maximum(-1, 29/2),Minimum(4/3 , 220/27)