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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?

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    BTW, to ensure the applicability of Rolle's Theorem, don't you want to check that your function is *differentiable*, not merely continuous? (Of course the differentiability of all polynomial functions is established early on in any calculus class.)2011-05-29

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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 $|f(x)g(x) - f(y)g(y)| = |f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)|$

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    Differentiability is needed here, not just continuity.2011-05-29