What is the critical x/(x^2+13x+36) ?

The critical x/(x^2+13x+36) is x=-6,x=6

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critical x/(x^2+13x+36)

critical points

\frac{x}{x ^{2} + 13 x + 36}

x = - 6, x = 6

Solution steps

\frac{x}{x ^{2} + 13 x + 36}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 6, x = 6, x = - 9, x = - 4

Identify critical points not in

f (x)

domain

Domain of

\frac{x}{x ^{2} + 13 x + 36} : x < - 9 or - 9 < x < - 4 or x > - 4

\frac{x}{x ^{2} + 13 x + 36}

is not defined at

x = - 9, x = - 4

, therefore:

x = - 6, x = 6

f (x) = \frac{( x - 1 ) ( x ^{2} + 4 )}{x ( x + 1 )}

56 n^{2} + 9 n + 9

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

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

4 x^{3} - 12 x^{2}