What is the extreme f(x)= 4/3 x^3+22x^2+96x+7 ?

The extreme f(x)= 4/3 x^3+22x^2+96x+7 is Maximum(-8,-107/3),Minimum(-3,-119)

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extreme f(x)= 4/3 x^3+22x^2+96x+7

extreme points

f (x) = \frac{4}{3} x^{3} + 22 x^{2} + 96 x + 7

Maximum (- 8, - \frac{107}{3}), Minimum (- 3, - 119)

Solution steps

f^{'} (x) = 4 x^{2} + 44 x + 96

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < - 3,

Increasing

: - 3 < x < \infty

Plug

x = - 8

into

\frac{4}{3} x^{3} + 22 x^{2} + 96 x + 7 : - \frac{107}{3}

Maximum (- 8, - \frac{107}{3})

Plug

x = - 3

into

\frac{4}{3} x^{3} + 22 x^{2} + 96 x + 7 : - 119

Minimum (- 3, - 119)

Maximum (- 8, - \frac{107}{3}), Minimum (- 3, - 119)

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

f (x, y) = 21 x y e^{- x^{2} - y^{2}}

2 x^{3} - 3 x^{2} - x + 9

f (x) = (x - 6) e^{- 6 x}

2 t^{10} (4 - t^{2})^{5}

What is the extreme f(x)= 4/3 x^3+22x^2+96x+7 ?
The extreme f(x)= 4/3 x^3+22x^2+96x+7 is Maximum(-8,-107/3),Minimum(-3,-119)