TITLE: If $f(x)≤x$ , then $f′(x)≤1$?
QUESTION [5 upvotes]: I'm studying Calculus and having a trouble solving this question.
1) If $f(x)\leq x$, then $f′(x)\leq 1$ for all $x$?
2) What if $f(0)=0$, $f′(x)$ exists for all $x$?
I could easily find the counter example for 1) (Therefore it is false)
But I'm not sure about 2)
If $f(0)=0$ and $f′(x)$ exists for all $x$  &
$f(x)\leq x$ , then $f′(x)\leq 1$ for all $x$?
Please leave a comment if you don't mind :)

REPLY [4 votes]: Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^{-x}$ has $f'(x)\gt1$ for all $x$, while $f(x)=x-{1\over2}x^2$ satisfies $f(0)=0$ but $f'(x)\gt1$ for $x\lt0$.
Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^{-x}$, of course, does not satisfy that condition.