GRE 9378 question 23 Let $f$ be a real valued function continuous on the closed interval $[0,1]$ and differentiable on the open interval $(0,1)$, with $f(0)=1$ and $f(1)=0$. Which of the following must be true?


*

*There exists $x \in (0,1)$ such that $f(x)=x$.

*There exists $x \in (0,1)$ such that $f'(x)=-1$.

*$f(x)>0 $ for all  $x \in [0,1)$


I think the solution is 1 only, but apparently is 1 and 2 only.
I don't see why 2 is true, if the derivative of $f$ is not necessarily continuous.
 A: This is a direct application of the mean value theorem, which does not require $f'$ to be continuous. 
A: Let $g(x)=f(x)+x$. Note that $g(0)=g(1)=1$.
If $g(x)$ is identically equal to $1$, there is nothing to prove. 
If $g(x)$ is not identically equal to $1$, then $g$ attains a local maximum or minimum in $(0,1)$. There we have $g'(x)=0$.
Remark: Even MVT was not used, though admittedly the crucial component of the proof of MVT was. 
A: The derivative may not be continuous, but it does have the Intermediate Value Property. 
A: Direct application of mean value theorem.
It is differentiable in open interval $(0, 1)$. Hence it is continuous too. So no matter what the graph is, it will have some point where the slope would be $-1$.
From $x=0$ to $x=1$, you have to connect $(0, 1)$ (as f(0)=1) and $(1, 0)$ (as f(1)=0).
You can consider some cases to do so.
Case 1:
A curve first going up, then coming down. So while coming down, it will have -1 slope somewhere
Case 2
A curve first going down, then coming up. While going down, it will have -1 slope somewhere
Case 3
A straight line connect the two points. Clearly, it has a slope of -1
So, 2$^{nd}$ must be true.
There can be infinite curves. I've explained just three.
While doing the question, if you can't figure it out, just try to search for a contradicting example.
