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!