What is the extreme y= 1/3 x^3+2x^2+3x ?

The extreme y= 1/3 x^3+2x^2+3x is Maximum(-3,0),Minimum(-1,-4/3)

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extreme y= 1/3 x^3+2x^2+3x

extreme

Maximum (- 3, 0), Minimum (- 1, - \frac{4}{3})

Solution steps

f^{'} (x) = x^{2} + 4 x + 3

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 3

into

\frac{1}{3} x^{3} + 2 x^{2} + 3 x : 0

Maximum (- 3, 0)

Plug

x = - 1

into

\frac{1}{3} x^{3} + 2 x^{2} + 3 x : - \frac{4}{3}

Minimum (- 1, - \frac{4}{3})

Maximum (- 3, 0), Minimum (- 1, - \frac{4}{3})

What is the extreme y= 1/3 x^3+2x^2+3x ?
The extreme y= 1/3 x^3+2x^2+3x is Maximum(-3,0),Minimum(-1,-4/3)