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

The extreme f(x)=(x^3)/3+(3x^2)/2 is Maximum(-3, 9/2),Minimum(0,0)

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

extreme points

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

Maximum (- 3, \frac{9}{2}), Minimum (0, 0)

Solution steps

f^{'} (x) = x^{2} + 3 x

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

\frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} : \frac{9}{2}

Maximum (- 3, \frac{9}{2})

Plug

x = 0

into

\frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} : 0

Minimum (0, 0)

Maximum (- 3, \frac{9}{2}), Minimum (0, 0)

2^{- x} - 4

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

- x - 2

3 x + 6 y = 12, (3, 3)

x^{2} - 3 x + 3

What is the extreme f(x)=(x^3)/3+(3x^2)/2 ?
The extreme f(x)=(x^3)/3+(3x^2)/2 is Maximum(-3, 9/2),Minimum(0,0)