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

The critical f(x)=x^3-3x^2-9x+4 is x=-1,x=3

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

critical points

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

x = - 1, x = 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 3

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - 1, x = 3

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

- \frac{4}{( x + 1 ) ^{2}}

4 x = 7 y + 5

f (x) = \frac{x + 6}{1 - x}

3 x - 2 y = 8, (4, - 2)

What is the critical f(x)=x^3-3x^2-9x+4 ?
The critical f(x)=x^3-3x^2-9x+4 is x=-1,x=3