$f$ is differentiable on $[a,b]$, $f&#39;(x) \leq A|f(x)|$ where $A$ is a non-negative constant.

If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$

I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.

I know $|f(x)|=|f&#39;(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\  |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.


Question: Password sending through network
Answer: SSL's only purpose is to stop Man in the middle from happening. Not having SSL/TLS just means you can't have:


Confidentiality
Integrity


Furthermore the password is just a token used for authentication, it's not required to have the actual password if the access token is the hash. You can perform a repeater/replay attack if you can sniff the hash from the network and provide it to your application to authenticate to your service.

An example for such an attack can be found in Windows. It's called the pass-the-hash attack.