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I am dealing with a set of functions with finitely many zeros. Are functions without any zeros included in that set, for example $f(x) = c \ne 0$?

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    I think that you can say "this function has $0$ zeros, and $0 \lt \infty$ so I'd say it has a finite number of zeroes – 2017-01-31
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    Sure, unless it's also specified that they need to have zeroes. – 2017-01-31
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    I don't like the way this question is worded. It's not about functions or roots - I'm sure you can tell that; it's barely related to set theory and more just a question about definitions whose best answer can only be 'what is the question exactly?' – 2017-03-15

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Yes, as the empty set is finite.

Edit: A common definition of a set $A$ to be finite is that there exists a injection $A\to\mathbb{N}$ but no surjection $A\to\mathbb{N}$. The empty set certainly satisfies this, so it is finite.

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They certainly are.

Let $f $ be a function with no zeroes. Let $A $ be the set of all $x$ such that $f(x)=0$. $A = \emptyset $, thus the cardinality of $A $ is $0$. $0$ is a finite number so $f $ is in your set.