GRE 9378 question 23

Question

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?

1. There exists $x \in (0,1)$ such that $f(x)=x$.
2. There exists $x \in (0,1)$ such that $f'(x)=-1$.
3. $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.

Answer 1

This is a direct application of the mean value theorem, which does not require $f'$ to be continuous.

Answer 2

The derivative may not be continuous, but it does have the Intermediate Value Property.

Answer 3

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.

Answer 4

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.