What is the domain of the function #f(x)=(3x^2)/(x^2-49)#?

1 Answer
Jan 23, 2015

The domain of a function is the set of the values in which you can calculate the function itself.

The problem of finding the domain of a functions is due to the fact that not every function "accepts" every real number as an input.

If you have a rational function, you must exclude from the domain the values for which the denominator is zero, since you cannot divide by zero. So, for example, if you consider the function #f(x)=\frac{1}{x-2}#, you see that you can evaluate it for every real value #x#, as long as it is not 2: in that case, you would have #f(2)=\frac{1}{2-2}#, an obviously invald operation. So, we say that the domain of #f# is the whole real number set, deprived of the element "2".

In your case, the denominator is the function #x^2-49#. Let's see for which values it is zero:
#x^2-49=0 \iff x^2=49 \iff x=\pm\sqrt{49} \iff x=\pm 7#.

These are the only two values you have to consider, since for any other #x\ne \pm 7#, the denominator isn't zero, and you have no problem calculating #f(x)#.

With a proper notation, your domain is the set #{x \in \mathbb{R} | x\ne\pm7}#