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 $f(x)\leqslant g(x)\leqslant h(x)$ 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.

3. Originally Posted by Drexel28
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 $f(x_0) = h(x_0)$

4. Originally Posted by southprkfan1
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

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'( $x_0$)=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)= $x^2$sin(1/x), x not equal to 0 and F(0)=0 has derivative equal to 0 about x=0?