What is the critical x^4+4/3 x^3-4x^2-4/3 ?

The critical x^4+4/3 x^3-4x^2-4/3 is x=-2,x=0,x=1

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critical x^4+4/3 x^3-4x^2-4/3

critical points

x^{4} + \frac{4}{3} x^{3} - 4 x^{2} - \frac{4}{3}

x = - 2, x = 0, x = 1

Solution steps

x^{4} + \frac{4}{3} x^{3} - 4 x^{2} - \frac{4}{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2, x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - 2, x = 0, x = 1

\frac{6 x}{1 - 7 x}

cos (2 x + 5)

y = 2 x + 1

f (x) = \frac{- 10}{10 - x}

f (x) = 64 - x^{2}

What is the critical x^4+4/3 x^3-4x^2-4/3 ?
The critical x^4+4/3 x^3-4x^2-4/3 is x=-2,x=0,x=1