How do you find the domain of #y = (3x^2)/(x^2+1)#?

1 Answer
Jun 13, 2015

Answer:

The domain is all the Real #x#.

Explanation:

The only problem with this function would be a value of #x# that makes it a division by zero; but the denominator will never become zero!
There is no Real value of #x# that makes #x^2+1# equal to zero.

#x^2+1=0# or #x^2=-1# NEVER

So this function can accept all the Real #x#.