What is the critical f(x)=x^3-4x^2+x+6 ?

The critical f(x)=x^3-4x^2+x+6 is x=(4-sqrt(13))/3 ,x=(4+sqrt(13))/3

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critical f(x)=x^3-4x^2+x+6

critical points

f (x) = x^{3} - 4 x^{2} + x + 6

x = \frac{4 - 13}{3}, x = \frac{4 + 13}{3}

Solution steps

x^{3} - 4 x^{2} + x + 6

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{4 - 13}{3}, x = \frac{4 + 13}{3}

Identify critical points not in

f (x)

domain

Domain of

x^{3} - 4 x^{2} + x + 6 : - \infty < x < \infty

All critical points are in domain

x = \frac{4 - 13}{3}, x = \frac{4 + 13}{3}

f (x) = \frac{x}{6 - x}

y = cos (x - \frac{π}{2})

f (x) = - 2 x^{4} + 7 x^{2}

\frac{x + 4}{2 x}

f (x) = \frac{x ^{2}}{2 x - 1}

What is the critical f(x)=x^3-4x^2+x+6 ?
The critical f(x)=x^3-4x^2+x+6 is x=(4-sqrt(13))/3 ,x=(4+sqrt(13))/3