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(-2, 13/3),Minimum(1,-1/6)

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

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

Maximum (- 2, \frac{13}{3})

Plug

x = 1

into

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

Minimum (1, - \frac{1}{6})

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

f (x) = y^{2} + x y - 2 x - 2 y + 2

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

f (x) = x^{4} - 5 x^{3} + 9 x^{2}

f (x, y) = x y - 2 x - y + 6

f (x) = - (x - 1)^{2}

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(-2, 13/3),Minimum(1,-1/6)