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Math Help - Differentiable question

  1. #1
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    Differentiable question

    Suppose that f,g,h, are real valued func on(a,b) and x0 belongs to (a,b). assume that f and h are differentiable at x0 and f(x0)=h(x0). f(x)<=g(x)<=h(x) for all x in (a,b). show that g is also differentiable and f'=g'=h' at x0.

    What i got so far is that by squeeze thm, we can get f(x0)=g(x0)=h(x0). and since f and h are differentiable at x0. then we have f(x)-f(x0)<=g(x)-g(x0)<=h(x)-h(x0). after that, we can conclude that g' exists. but i have no idea of how could i show f'=g'=h'. Please help! Thanks!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by shuxue View Post
    Suppose that f,g,h, are real valued func on(a,b) and x0 belongs to (a,b). assume that f and h are differentiable at x0 and f(x0)=h(x0). f(x)<=g(x)<=h(x) for all x in (a,b). show that g is also differentiable and f'=g'=h' at x0.

    What i got so far is that by squeeze thm, we can get f(x0)=g(x0)=h(x0). and since f and h are differentiable at x0. then we have f(x)-f(x0)<=g(x)-g(x0)<=h(x)-h(x0). after that, we can conclude that g' exists. but i have no idea of how could i show f'=g'=h'. Please help! Thanks!
    We have that f(x)\leqslant g(x)\leqslant h(x) and that \lim_{x\to x_0}f\frac{(x)-f(x_0)}{x-x_0}=f'(x_0)\leqslant\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}\leqslant \lim_{x\to x_0}\frac{h(x)-h(x_0)}{x-x_0}=h'(x_0)=f'(x_0) and so the conclusion follows by the squeeze theorem. What's the problem? You were right.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    We have that f(x)\leqslant g(x)\leqslant h(x) and that \lim_{x\to x_0}f\frac{(x)-f(x_0)}{x-x_0}=f'(x_0)\leqslant\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}\leqslant \lim_{x\to x_0}\frac{h(x)-h(x_0)}{x-x_0}=h'(x_0)=f'(x_0) and so the conclusion follows by the squeeze theorem. What's the problem? You were right.
    Drexel28, I was thinking something along those lines too. But, how do you know  f'(x_0) = h'(x_0) ? All he says is that  f(x_0) = h(x_0)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Drexel28, I was thinking something along those lines too. But, how do you know  f'(x_0) = h'(x_0) ? All he says is that  f(x_0) = h(x_0)
    Editing
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Editing
    so I have like the same question lol. But I'm also not understanding how to show or how you know that f'( x_0)=g'( x_0)=h'( x_0)...IDK...and also how would you use this result to explain why the function F(x) defined by F(x)= x^2sin(1/x), x not equal to 0 and F(0)=0 has derivative equal to 0 about x=0?
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