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

The extreme f(x)=(x^3)/3-(x^2)/2-2x+1 is Maximum(-1, 13/6),Minimum(2,-7/3)

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

extreme

Maximum (- 1, \frac{13}{6}), Minimum (2, - \frac{7}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x + 1 : \frac{13}{6}

Maximum (- 1, \frac{13}{6})

Plug

x = 2

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x + 1 : - \frac{7}{3}

Minimum (2, - \frac{7}{3})

Maximum (- 1, \frac{13}{6}), Minimum (2, - \frac{7}{3})

What is the extreme f(x)=(x^3)/3-(x^2)/2-2x+1 ?
The extreme f(x)=(x^3)/3-(x^2)/2-2x+1 is Maximum(-1, 13/6),Minimum(2,-7/3)