What is the extreme f(x)=(x^3)/3-(x^2)/2-6x+4 ?

The extreme f(x)=(x^3)/3-(x^2)/2-6x+4 is Maximum(-2, 34/3),Minimum(3,-19/2)

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extreme f(x)=(x^3)/3-(x^2)/2-6x+4

extreme points

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

Maximum (- 2, \frac{34}{3}), Minimum (3, - \frac{19}{2})

Solution steps

f^{'} (x) = x^{2} - x - 6

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 6 x + 4 : \frac{34}{3}

Maximum (- 2, \frac{34}{3})

Plug

x = 3

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 6 x + 4 : - \frac{19}{2}

Minimum (3, - \frac{19}{2})

Maximum (- 2, \frac{34}{3}), Minimum (3, - \frac{19}{2})

What is the extreme f(x)=(x^3)/3-(x^2)/2-6x+4 ?
The extreme f(x)=(x^3)/3-(x^2)/2-6x+4 is Maximum(-2, 34/3),Minimum(3,-19/2)