1. ## 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!

2. Originally Posted by shuxue
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 $\displaystyle f(x)\leqslant g(x)\leqslant h(x)$ and that $\displaystyle \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$$\displaystyle \lim_{x\to x_0}\frac{h(x)-h(x_0)}{x-x_0}=h'(x_0)=f'(x_0) and so the conclusion follows by the squeeze theorem. What's the problem? You were right. 3. Originally Posted by Drexel28 We have that \displaystyle f(x)\leqslant g(x)\leqslant h(x) and that \displaystyle \lim_{x\to x_0}f\frac{(x)-f(x_0)}{x-x_0}=f'(x_0)\leqslant\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}\leqslant$$\displaystyle \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 $\displaystyle f'(x_0) = h'(x_0)$? All he says is that $\displaystyle f(x_0) = h(x_0)$

4. Originally Posted by southprkfan1
Drexel28, I was thinking something along those lines too. But, how do you know $\displaystyle f'(x_0) = h'(x_0)$? All he says is that $\displaystyle f(x_0) = h(x_0)$
Editing

5. Originally Posted by Drexel28
Editing
so I have like the same question lol. But I'm also not understanding how to show or how you know that f'($\displaystyle x_0$)=g'($\displaystyle x_0$)=h'($\displaystyle x_0$)...IDK...and also how would you use this result to explain why the function 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 about x=0?