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

The critical f(x)=x^4-3x^3+3x^2+1 is x=0

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critical f(x)=x^4-3x^3+3x^2+1

critical points

f (x) = x^{4} - 3 x^{3} + 3 x^{2} + 1

x = 0

Solution steps

x^{4} - 3 x^{3} + 3 x^{2} + 1

Find where

f^{'} (x)

is equal to zero or undefined

x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 3 x^{3} + 3 x^{2} + 1 : - \infty < x < \infty

All critical points are in domain

x = 0

(0, 8), (24.08, 0)

- 2 + 3 sin (2 x + \frac{π}{4})

f (x) = (- 5 x + 1)^{2}

f (x) = \frac{2 x}{x ^{2} - 9}

4 cos (\frac{1}{3} x + \frac{π}{4}) + 1

What is the critical f(x)=x^4-3x^3+3x^2+1 ?
The critical f(x)=x^4-3x^3+3x^2+1 is x=0