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

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

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

critical points

\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

y = 2 x - x^{2}

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

f (x, y) = 3 x^{2} + x^{2} y + 6 y^{2} + y^{3}

f (x) = x^{4} - 32 x + 2

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

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