Question

What is the domain of `f(x)=x/(x^2-5x)`?

Answer 1

`D = -oo < x < oo | x != 0, x!= 5 and x in RR`

Explanation:

The domain is every value that `x` can take without having a math error (division by zero, logarithm of a null or negative number, even root of a negative number, etc.)

So the only caveat we have here is that the denominator mustn't be 0. Or
`x^2 - 5x != 0`

We can solve this using the quadratic formula, sum and product, or, just do the easy thing and factor it out.
`x^2 - 5x != 0`
`x(x - 5) != 0`

Since the product can't be zero, neither can, that is
`x != 0`
`x - 5 != 0 rarr x != 5`

So the domain D, is `D = -oo < x < oo, x != 0, x!= 5 | x in RR`
Or
`D = -oo < x < 0 or 0 < x < 5 or 5 < x | x in RR`
Or that same thing in set notation.