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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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    Do you know how to show that the sum and product of continuous (real-valued) functions is continuous?2011-05-29
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    @Qiaochu Yuan Hmm, you mean breaking $f$ into smaller elementary functions (is that how they're called in English?) like $g(x) = x-2$ and so on? My teacher merely told me that such functions don't need to be proven to be continuous: you only have to mention that they are elementary.2011-05-29
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    Well, what you do and don't need to prove continuous is a function of your teacher's expectations, not a mathematical fact.2011-05-29
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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