What is the extreme f(x)= 1/3 x^3+3x^2+8x ?

The extreme f(x)= 1/3 x^3+3x^2+8x is Maximum(-4,-16/3),Minimum(-2,-20/3)

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extreme f(x)= 1/3 x^3+3x^2+8x

extreme points

f (x) = \frac{1}{3} x^{3} + 3 x^{2} + 8 x

Maximum (- 4, - \frac{16}{3}), Minimum (- 2, - \frac{20}{3})

Solution steps

f^{'} (x) = x^{2} + 6 x + 8

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 4

into

\frac{1}{3} x^{3} + 3 x^{2} + 8 x : - \frac{16}{3}

Maximum (- 4, - \frac{16}{3})

Plug

x = - 2

into

\frac{1}{3} x^{3} + 3 x^{2} + 8 x : - \frac{20}{3}

Minimum (- 2, - \frac{20}{3})

Maximum (- 4, - \frac{16}{3}), Minimum (- 2, - \frac{20}{3})

m = - 2, (1, 0)

(0, 0), (1, 3)

f (x) = \frac{x ^{2}}{x - 5}

f (x) = - 2 x^{3} - 7

What is the extreme f(x)= 1/3 x^3+3x^2+8x ?
The extreme f(x)= 1/3 x^3+3x^2+8x is Maximum(-4,-16/3),Minimum(-2,-20/3)