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

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

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

critical points

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

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

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

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

- x^{3} - 6 x^{2} + 3 x + 18

f (x) = 2 x^{3} + 3 x^{2} - 12 x

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

\frac{x ^{2} + 1}{x ^{2} - x - 2}

f (x) = x^{2} - 5 x - 24

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