What is the critical f(x)=x^3+5x^2-4x+1 ?

The critical f(x)=x^3+5x^2-4x+1 is x=-(5+sqrt(37))/3 ,x=(-5+sqrt(37))/3

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critical f(x)=x^3+5x^2-4x+1

critical points

f (x) = x^{3} + 5 x^{2} - 4 x + 1

x = - \frac{5 + 37}{3}, x = \frac{- 5 + 37}{3}

Solution steps

x^{3} + 5 x^{2} - 4 x + 1

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{5 + 37}{3}, x = \frac{- 5 + 37}{3}

Identify critical points not in

f (x)

domain

Domain of

x^{3} + 5 x^{2} - 4 x + 1 : - \infty < x < \infty

All critical points are in domain

x = - \frac{5 + 37}{3}, x = \frac{- 5 + 37}{3}

(2, 4), (- 3, - 8)

f (t) = t

f (x) = ln (4 - x)

f (x) = 3 sin (2 x)

f (x) = x + 3

What is the critical f(x)=x^3+5x^2-4x+1 ?
The critical f(x)=x^3+5x^2-4x+1 is x=-(5+sqrt(37))/3 ,x=(-5+sqrt(37))/3