What is the extreme f(x)= 8/3 x^3+32x^2+120x+9 ?

The extreme f(x)= 8/3 x^3+32x^2+120x+9 is Maximum(-5,-373/3),Minimum(-3,-135)

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extreme f(x)= 8/3 x^3+32x^2+120x+9

extreme

Maximum (- 5, - \frac{373}{3}), Minimum (- 3, - 135)

Solution steps

f^{'} (x) = 8 x^{2} + 64 x + 120

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 3,

Increasing

: - 3 < x < \infty

Plug

x = - 5

into

\frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9 : - \frac{373}{3}

Maximum (- 5, - \frac{373}{3})

Plug

x = - 3

into

\frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9 : - 135

Minimum (- 3, - 135)

Maximum (- 5, - \frac{373}{3}), Minimum (- 3, - 135)

What is the extreme f(x)= 8/3 x^3+32x^2+120x+9 ?
The extreme f(x)= 8/3 x^3+32x^2+120x+9 is Maximum(-5,-373/3),Minimum(-3,-135)