How do you find the domain of #g(x)= x/(x^2-5x)#?

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Answer 1 · 2017-10-25T16:03:27

#x in RR, x != 0,5#

The domain is the range of #x# values a function can take.

The numerator of #g(x)# is simply #x# so it doesn't really matter what it is.

The denominator is more complex because we know that we can never divide by #0#.

If the denominator is #0#, then

#x^2 - 5x = 0#

#x(x-5) = 0#

so

#x = 0, 5#

In which case the domain is #x in RR, x != 0, 5#