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

1 Answer
Sep 19, 2015

Answer:

#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.