What is the critical f(x)=4x^3-33x^2-36x+2 ?

The critical f(x)=4x^3-33x^2-36x+2 is x=-1/2 ,x=6

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critical f(x)=4x^3-33x^2-36x+2

critical points

f (x) = 4 x^{3} - 33 x^{2} - 36 x + 2

x = - \frac{1}{2}, x = 6

Solution steps

4 x^{3} - 33 x^{2} - 36 x + 2

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{2}, x = 6

Identify critical points not in

f (x)

domain

Domain of

4 x^{3} - 33 x^{2} - 36 x + 2 : - \infty < x < \infty

All critical points are in domain

x = - \frac{1}{2}, x = 6

\frac{1}{x + 3}

f (x) = \frac{x ^{3} - 7}{2}

4 x^{2}, x \leq 0

(0, 1), (1, 0)

(7.14) (- 1.9)

What is the critical f(x)=4x^3-33x^2-36x+2 ?
The critical f(x)=4x^3-33x^2-36x+2 is x=-1/2 ,x=6