## Differentiability

Suppose that f : [a, b] to R is differentiable and c is an element of [a, b]. Then show that there exists a sequence {x_n} converges to c where x_n does not equal c for all n, such that f '(c) = lim f ' (x_n) Note: I use the definition of differentiation of: lim (x goes to c) (f(x) - f(c)) / (x - c).

• Let h(x) = f(x) - g(x). Note that h(a) = f(a) - g(a) = 0.
It is easy to check that h is continuous on [a,b] and differentiable on (a,b).

Thus the Mean Value Theorem applies, and we get
[h(b) - h(a)] / (b - a) = h'(c) for some c in (a,b).
==> [(f(b) - g(b)) - 0] / (b - a) = f'(c) - g'(c) < 0, since f '(x)<g '(x) for a<x<b.

Thus, [(f(b) - g(b)) - 0] / (b - a) < 0
==> f(b) - g(b) < 0
==> f(b) < g(b), as required.