What is the extreme f(x)=x^3-6x^2-96x ?

The extreme f(x)=x^3-6x^2-96x is Maximum(-4,224),Minimum(8,-640)

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extreme f(x)=x^3-6x^2-96x

extreme points

f (x) = x^{3} - 6 x^{2} - 96 x

Maximum (- 4, 224), Minimum (8, - 640)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x - 96

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = - 4

into

x^{3} - 6 x^{2} - 96 x : 224

Maximum (- 4, 224)

Plug

x = 8

into

x^{3} - 6 x^{2} - 96 x : - 640

Minimum (8, - 640)

Maximum (- 4, 224), Minimum (8, - 640)

f (x) = x + 4 - 2

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

y = - (X - 3)^{2} + 1

f (x) = x^{\frac{3}{4}}

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

What is the extreme f(x)=x^3-6x^2-96x ?
The extreme f(x)=x^3-6x^2-96x is Maximum(-4,224),Minimum(8,-640)