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Thread: show that g is differentiable at x_0 and that f'(x_0)=g'(x_0)=h'(x_0)...?

  1. #1
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    show that g is differentiable at x_0 and that f'(x_0)=g'(x_0)=h'(x_0)...?

    Suppose that f,g,h are real valued functions on (a,b) and $\displaystyle x_0$$\displaystyle \in$(a,b). Assume that f and h are differentiable at $\displaystyle x_0$, f($\displaystyle x_0$)=h($\displaystyle x_0$) and f(x)<=g(x)<=h(x) for all x$\displaystyle \in$(a,b). show that g is differentiable at $\displaystyle x_0$ and f'($\displaystyle x_0$)=g'($\displaystyle x_0$)=h'($\displaystyle x_0$). Use this result to explain why the fraction F(x) defined by F(x)=$\displaystyle x^2$sin(1/x), x not equal to 0, and F(0)=0 has derivative equal to 0 at x=0.

    I honestly have no clue I hope some one can help me thanks!
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  2. #2
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    Quote Originally Posted by alice8675309 View Post
    Suppose that f,g,h are real valued functions on (a,b) and $\displaystyle x_0$$\displaystyle \in$(a,b). Assume that f and h are differentiable at $\displaystyle x_0$, f($\displaystyle x_0$)=h($\displaystyle x_0$) and f(x)<=g(x)<=h(x) for all x$\displaystyle \in$(a,b). show that g is differentiable at $\displaystyle x_0$ and f'($\displaystyle x_0$)=g'($\displaystyle x_0$)=h'($\displaystyle x_0$). Use this result to explain why the fraction F(x) defined by F(x)=$\displaystyle x^2$sin(1/x), x not equal to 0, and F(0)=0 has derivative equal to 0 at x=0.

    I honestly have no clue I hope some one can help me thanks!
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  3. #3
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    Quote Originally Posted by southprkfan1 View Post
    I've noticed that this question was already posted (sadly after I posted) however I haven't a clue who that person is lol. I find that there are usually similar questions being posted which is a plus b/c its good practice. But I mean so many school use the same or similar books so I doubt it. The one I posted had extra parts to it also. That'd be cool tho and random lol
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