# Prove

• October 22nd 2012, 08:58 AM
franios
Prove
Suppose f,g, & h are defined on (a,b) & a<x0<b. Assume f and h are differentiable at x0, f(x0)=h(x0), & f(x)<g(x)<h(x) for all x in (a,b). Prove that g is differentiable at x0 and f'(x0)=g'(x0)=g'(x0).
• October 22nd 2012, 09:33 AM
Plato
Re: Prove
Quote:

Originally Posted by franios
Suppose f,g, & h are defined on (a,b) & a<x0<b. Assume f and h are differentiable at x0, f(x0)=h(x0), & f(x)<g(x)<h(x) for all x in (a,b). Prove that g is differentiable at x0 and f'(x0)=g'(x0)=g'(x0).

It is important that $f(x_0)=h(x_0)$ so $-f(x_0)=-h(x_0)$.

If $\delta>0$ can you show that
$\frac{f(x_0+\delta)-f(x_0)}{\delta}<\frac{g(x_0+\delta)-g(x_0)}{\delta}<\frac{h(x_0+\delta)-h(x_0)}{\delta}~?$

There is a similar statement if $\delta<0$.

Those are used on the limit definition of the derivatives and $f'(x_0)=h'(x_0)$.