What is the critical f(x)=x^4+4x^3-8x^2+1 ?

The critical f(x)=x^4+4x^3-8x^2+1 is x=-4,x=0,x=1

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critical f(x)=x^4+4x^3-8x^2+1

critical points

f (x) = x^{4} + 4 x^{3} - 8 x^{2} + 1

x = - 4, x = 0, x = 1

Solution steps

x^{4} + 4 x^{3} - 8 x^{2} + 1

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

x^{4} + 4 x^{3} - 8 x^{2} + 1 : - \infty < x < \infty

All critical points are in domain

x = - 4, x = 0, x = 1

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

ln (\frac{1}{x - 1} + 1)

f (x) = \frac{x + 5}{x - 8}

(70.3) (90.2)

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

What is the critical f(x)=x^4+4x^3-8x^2+1 ?
The critical f(x)=x^4+4x^3-8x^2+1 is x=-4,x=0,x=1