# Positive derivative implies increasing without Mean Value Theorem

The result below is usually proven by using the Mean Value Theorem (see e.g. ProofWiki).

But can we also prove it more directly (and fairly elegantly) without resort to the MVT?

Suppose $$f:[a,b]\rightarrow \mathbb{R}$$ is differentiable.

• If $$f'(x)\geq0$$ for all $$x\in[a,b]$$, then $$f$$ is increasing on $$[a,b]$$.
• If $$f'(x)>0$$ for all $$x\in[a,b]$$, then $$f$$ is strictly increasing on $$[a,b]$$.
• If $$f'(x)\leq0$$ for all $$x\in[a,b]$$, then $$f$$ is decreasing on $$[a,b]$$.
• If $$f'(x)<0$$ for all $$x\in[a,b]$$, then $$f$$ is strictly decreasing on $$[a,b]$$.
• The MVT is at the heart of the matter (and the MVT-based proof is elegant). – quasi Aug 8 '19 at 2:19
• @quasi: I do not disagree with you. But would you also say that it is impossible to prove this result without using the MVT (or ideas involved in the MVT)? – iqntt1s Aug 8 '19 at 2:29
• I doubt that you can do it more simply or more elegantly than the standard MVT-based proof. Moreover, any alternative proof would likely use the MVT in camouflage. – quasi Aug 8 '19 at 2:30
• Your statement plus Darboux's theorem implies Rolle's theorem and hence MVT pretty directly, which seems like a strong indication that this can't be proved without essentially repeating a proof of MVT. – Nate Eldredge Aug 8 '19 at 2:38
• Would you be interested in proofs for open intervals $(a,b)$ rather than closed? – coffeemath Aug 8 '19 at 2:38

## 1 Answer

Long comment: If you want to learn to prove things about differentiable functions then this is the wrong question! Because MVT is the way things about differentiable functions are proved.

Of course there must exist counterexamples to that last statement, but regardless you'll be better off if, when you need to prove something about diifferentiable functions, you automatically consider whether you can apply MVT.

If you want to master elementary-calculus-with-proofs you should

1. Study the proof of Rolle's theorem until it seems "obvious".

2. Similarly for the proof that Rolle implies MVT.

3. Similarly for the proof of the current result using MVT.

Honest. If, as seems possible, I'm better at proving things about derivatives than you are, the reason is I did (1), (2) and (3) long ago. Wondering how you can avoid MVT here is not going to be nearly as useful.