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?