# Why are so many people under the impression that we need to find the domain of a rational function in order to find its zeros? Zeros of f(x) = (x^2-x)/(3x^4+4x^3-7x+9) are 0,1.

May 18, 2015

I think that finding the domain of a rational function isn't necessarily related to finding its roots/zeros. Finding the domain simply means finding the preconditions for the mere existence of the rational function.

In other words, before finding its roots, we have to make sure under what conditions the function does exist. It might seem pedantic to do so, but there are particular cases when this matters.

Jan 29, 2016

My guess is that a factor in the numerator could also be represented in the denominator, resulting in a removable discontinuity.

#### Explanation:

This is just my speculation, but I'd bet the problem occurs with finding the zeros of a function like this:

$\frac{{x}^{2} - 3 x}{{x}^{3} + 2 {x}^{2} - 29 x + 42}$

You'd be tempted to say the zeros are at $x = 0$ and $x = 3$, but really there's only a zero at $x = 0$.

If you factor the denominator (and numerator), you get

$\frac{x \left(x - 3\right)}{\left(x - 3\right) \left(x - 2\right) \left(x + 7\right)}$

So the function is really just $\frac{x}{\left(x - 2\right) \left(x + 7\right)}$ with a hole at $x = 3$.

Edit:

This also could apply to functions with odder denominators. I really don't think this is incredibly important to note, since it's rare this is ever an issue, but in

$\frac{1}{x \sin x}$

The domain doesn't include $x = 0 , \pi , 2 \pi \ldots$

So in a function like

$\frac{x - \pi}{x \sin x}$

There's not a zero at $x = \pi$ but just a hole. So, I could see the value in looking at the domain to make sure there's no overlap in domain restrictions and possible zeros for odder functions like this.