What is the critical x^3-11x^2+39x-47 ?

The critical x^3-11x^2+39x-47 is x=3,x= 13/3

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critical x^3-11x^2+39x-47

critical points

x^{3} - 11 x^{2} + 39 x - 47

x = 3, x = \frac{13}{3}

Solution steps

x^{3} - 11 x^{2} + 39 x - 47

Find where

f^{'} (x)

is equal to zero or undefined

x = 3, x = \frac{13}{3}

Identify critical points not in

f (x)

domain

Domain of

x^{3} - 11 x^{2} + 39 x - 47 : - \infty < x < \infty

All critical points are in domain

x = 3, x = \frac{13}{3}

8 y - 3 x + 6 = 0

\frac{6 x}{36 - x ^{2}}

f (x) = \frac{x - 7}{x + 2}

f (x) = e^{- ∣ x ∣}

\frac{19}{x ^{3}}

What is the critical x^3-11x^2+39x-47 ?
The critical x^3-11x^2+39x-47 is x=3,x= 13/3