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Math Help - Zero Derivative Theorem

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    Zero Derivative Theorem

    I need to prove the zero derivative theorem. The theorem states that if a differentiable function f defined on the closed interval [a,b] has for every x in (a,b) f has a zero derivative at x, then the function f is constant.
    I have an intuitive understanding of why this is true, but I'm not sure how to make this formal. Any help would be great. Thanks
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ion599 View Post
    I need to prove the zero derivative theorem. The theorem states that if a differentiable function f defined on the closed interval [a,b] has for every x in (a,b) f has a zero derivative at x, then the function f is constant.
    I have an intuitive understanding of why this is true, but I'm not sure how to make this formal. Any help would be great. Thanks
    Problem: Let f be a differentiable function defined on [a,b] such that f'(x)=0\quad\forall x\in(a,b) Prove that f is a constant function.

    Proof: Let (\varepsilon,\varepsilon_1) be some arbitrary subinterval of (a,b). By the mean value theorem we know that there exists some \delta\in(\varepsilon,\varepsilon_1) such that \frac{f(\varepsilon_1)-f(\varepsilon)}{\varepsilon_1-\varepsilon}=f'(\delta) but since \delta\in(a,b) we have by assumption that f'(\delta)=0. Therefore f(\varepsilon_1)-f(\varepsilon)=0. Since \varepsilon_1,\varepsilon were arbitrary the conclusion follows.

    Remark: In retrospect, all though I think the above is fine it may have been more intuitive to consider any arbitrary subinterval (a,\varepsilon)\subset (a,b) which would then show that f(a)=f(\varepsilon) for any arbitrary \varepsilon\in(a,b)
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  3. #3
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    Ok thank you that was very helpful.
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