TITLE: Continuity: Does this need to be proven?
QUESTION [5 upvotes]: Given a function $f(x) = (x-2)(x-3)(x-4)(x-5) + 1$, I am asked to show that $f'(x) = 0$ has exactly three distinct roots. This is simple enough, it's done with Rolle's theorem: $f(2) = f(3) = f(4) = f(5) = 1$ and from there on it's rather easy.
However, for Rolle's theorem to be applicable I have to show that the function is continuous on the $[2, 3], [3, 4]$ and $[4, 5]$ closed intervals. How can I do that? Is it obvious from the function definition?

REPLY [5 votes]: Note that when you multiply your $f$ you get a polynomial of degree $4$ and note that polynomials are differentiable functions.
Or if you want to prove the statement explicitly, then try using the $\epsilon - \delta$ definition of continuity with the fact that $$|f(x)g(x) - f(y)g(y)| = |f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)|$$