What is the critical f(x)=((x^2+x-38))/((x^2-25)) ?

The critical f(x)=((x^2+x-38))/((x^2-25)) is x=1,x=25

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critical f(x)=((x^2+x-38))/((x^2-25))

critical points

f (x) = \frac{( x ^{2} + x - 38 )}{( x ^{2} - 25 )}

x = 1, x = 25

Solution steps

\frac{x ^{2} + x - 38}{x ^{2} - 25}

Find where

f^{'} (x)

is equal to zero or undefined

x = 1, x = 25, x = - 5, x = 5

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} + x - 38}{x ^{2} - 25} : x < - 5 or - 5 < x < 5 or x > 5

\frac{x ^{2} + x - 38}{x ^{2} - 25}

is not defined at

x = - 5, x = 5

, therefore:

x = 1, x = 25

f (x) = (8 + x) (y - x)

f (x) = \frac{4 x}{3 ( x ^{2} - 4 ) ^{\frac{1}{3}}}

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

f (x) = x + \frac{36}{x}

f (x) = 9 x^{4} + 6 x^{3} + 1

What is the critical f(x)=((x^2+x-38))/((x^2-25)) ?
The critical f(x)=((x^2+x-38))/((x^2-25)) is x=1,x=25