What is the critical f(x)=4x^3+6x^2-72x-9 ?

The critical f(x)=4x^3+6x^2-72x-9 is x=-3,x=2

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critical f(x)=4x^3+6x^2-72x-9

critical points

f (x) = 4 x^{3} + 6 x^{2} - 72 x - 9

x = - 3, x = 2

Solution steps

4 x^{3} + 6 x^{2} - 72 x - 9

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 2

Identify critical points not in

f (x)

domain

Domain of

4 x^{3} + 6 x^{2} - 72 x - 9 : - \infty < x < \infty

All critical points are in domain

x = - 3, x = 2

x y^{2} + 10 = 0

lo g_{4} (x) + 2

f (x) = x^{2} + 8 x - 1

f (x) = 1 - 2^{t}

f (x) = x^{2} e^{x}

What is the critical f(x)=4x^3+6x^2-72x-9 ?
The critical f(x)=4x^3+6x^2-72x-9 is x=-3,x=2