What is the critical f(x)=8x^3-12x^2-48x ?

The critical f(x)=8x^3-12x^2-48x is x=-1,x=2

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critical f(x)=8x^3-12x^2-48x

critical points

f (x) = 8 x^{3} - 12 x^{2} - 48 x

x = - 1, x = 2

Solution steps

8 x^{3} - 12 x^{2} - 48 x

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 2

Identify critical points not in

f (x)

domain

Domain of

8 x^{3} - 12 x^{2} - 48 x : - \infty < x < \infty

All critical points are in domain

x = - 1, x = 2

f (x) = x + x + 6

y - 7 = 8 (x - 14)

2 x^{3} + 6 x^{2} + 2

f (x) = tan (x)

y = 3 x^{2} - 3

What is the critical f(x)=8x^3-12x^2-48x ?
The critical f(x)=8x^3-12x^2-48x is x=-1,x=2