What is the domain of y= 1/(x^2-5) ?

The domain of y= 1/(x^2-5) is x sqrt(5)

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domain of y= 1/(x^2-5)

domain

y = \frac{1}{x ^{2} - 5}

x < - 5 or - 5 < x < 5 or x > 5

Solution steps

Find undefined (singularity) points:

x = 5, x = - 5

The function domain

x < - 5 or - 5 < x < 5 or x > 5

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

- cos (2 x)

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

f (x) = lo g_{2} (x + 1) + lo g_{2} (x - 1)

f (x) = \frac{2 x}{x - 5}

What is the domain of y= 1/(x^2-5) ?
The domain of y= 1/(x^2-5) is x<-sqrt(5)\lor-sqrt(5)<x<sqrt(5)\lor x>sqrt(5)