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

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

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

critical points

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

x = 0

Solution steps

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

All critical points are in domain

x = 0

lo g_{3} (x - 1)

(5, - 7), (- 2, 4)

f (x) = \frac{8 x}{7 x - 3}

g (x) = \frac{x + 9}{x ^{2} - 16}

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