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

• Apr 21st 2010, 09:23 PM
alice8675309
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$$\displaystyle \in(a,b). Assume that f and h are differentiable at \displaystyle x_0, f(\displaystyle x_0)=h(\displaystyle x_0) and f(x)<=g(x)<=h(x) for all x\displaystyle \in(a,b). show that g is differentiable at \displaystyle x_0 and f'(\displaystyle x_0)=g'(\displaystyle x_0)=h'(\displaystyle x_0). Use this result to explain why the fraction F(x) defined by F(x)=\displaystyle x^2sin(1/x), x not equal to 0, and F(0)=0 has derivative equal to 0 at x=0. I honestly have no clue :( I hope some one can help me thanks! • Apr 22nd 2010, 04:47 AM southprkfan1 Quote: Originally Posted by alice8675309 Suppose that f,g,h are real valued functions on (a,b) and \displaystyle x_0$$\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!

Are you 2 in the same class? http://www.mathhelpforum.com/math-he...-question.html
• Apr 22nd 2010, 05:57 AM
alice8675309
Quote:

Originally Posted by southprkfan1

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