Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 10E / i

Table of contents
- Solution i

Solution

Because

f^{''}

exists everywhere,

f^{'}

is differentiable everywhere and therefore continuous everywhere; continuity of

f^{''}

is not required. For

x < y

, apply the mean value theorem to

f^{'}

on

[x, y]

. It gives

ξ \in (x, y)

with

f^{'} (y) - f^{'} (x) = (y - x) f^{''} (ξ) \geq 0.

For

x = y

equality is immediate. Hence $f^{'}$ is nondecreasing: $x \leq y$ implies $f^{'} (x) \leq f^{'} (y)$ .

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