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, y) = 7 + 6 x^{4} + 5 y^{4}

f (x) = x^{3} - 3 x y + y^{3}

h (t) = t^{\frac{3}{4}} - 2 t^{\frac{1}{4}}

f (x) = 8 x^{3} - x^{2}

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

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