What is the extreme f(x)=(x^3)/3-(x^2)/2-2x-9 ?

The extreme f(x)=(x^3)/3-(x^2)/2-2x-9 is Maximum(-1,-47/6),Minimum(2,-37/3)

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extreme f(x)=(x^3)/3-(x^2)/2-2x-9

extreme points

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x - 9

Maximum (- 1, - \frac{47}{6}), Minimum (2, - \frac{37}{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 - 9 : - \frac{47}{6}

Maximum (- 1, - \frac{47}{6})

Plug

x = 2

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x - 9 : - \frac{37}{3}

Minimum (2, - \frac{37}{3})

Maximum (- 1, - \frac{47}{6}), Minimum (2, - \frac{37}{3})

f (x) = x^{2} (2 x - 1)^{\frac{1}{3}} [0.1]

f (x) = (sin (x) + cos (x))^{2}

\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8

f (x) = - 4 (x - 1)^{\frac{4}{7}} + 8

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

What is the extreme f(x)=(x^3)/3-(x^2)/2-2x-9 ?
The extreme f(x)=(x^3)/3-(x^2)/2-2x-9 is Maximum(-1,-47/6),Minimum(2,-37/3)