**Problem Statement:** Suppose satisfies for every and for a constant. Prove is a constant function.

**Proof: **Let . Then the absolute value of the derivative is defined to be . By our assumption it follows that

Since it follows that , thus:

Since this limit exists for every we may conclude that is differentiable on . Furthermore, we may conclude that for all which implies that for all , thus, for all . Therefore, is a constant function on .

**Reflection: **This is one of those proofs that really falls straight from the definitions and the hypotheses. If we didn’t have that then we couldn’t conclude that we didn’t have a zero in the denominator. This one was very straightforward and clean.

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