How many solutions does $f'(x)=1/2$ have, given that $|f(x)-f(y)|\ge|x-y|$? 
Let $F: \mathbb R \to \mathbb R$ be a continuously differentiable function such that
$$|f(x)-f(y)| \ge |x-y|,\ \  \forall x,y \in \mathbb R$$
Then $ f '(x)=1/2$ has how many solutions?

The way I approached this problem was like this,
$$\frac{|f(x) - f(y)|}{|x-y|} \ge 1.$$
So this implies that the slope of this function is either $1$ or more than $1$, over $\mathbb R$.
And the differential equation given has the solution
$$y = x/2 + c$$
implying that it has a slope of $1/2$.
Is it wrong that I am taking the given function and the solution of the differential equation as two distinct graphs and then looking at their solution?
Edit: As it turns out, I am indeed wrong. This was a much simpler problem and I complicated it. I cannot take 2 different functions because they are talking about the same function.
Thanks for the comments and replies.
 A: As with many things in analysis, it boils down to using the triangle inequality and referring back to the definitions.
We know for all $\epsilon>0$ there is some $\delta$ such that $|x-y|<\delta$ implies $\left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|<\epsilon$. So now we can start to set up our triangle inequality,
$$\left|\frac{f(x)-f(y)}{x-y}\right| = \left|\left(\frac{f(x)-f(y)}{x-y}-f'(x)\right)+f'(x)\right| \le \left|\frac{f(x)-f(y)}{x-y}-f'(x)\right| +|f'(x)|$$
So let's pick $\epsilon = \frac{1}{2}$ and plug in $|f'(x)|=\frac{1}{2}$ and we have ultimately that,
$$\left|\frac{f(x)-f(y)}{x-y}\right| <\frac{1}{2} + \frac{1}{2} = 1$$
However you've already shown that $\left|\frac{f(x)-f(y)}{x-y}\right|\ge 1$ which is in direct contradiction, so there can be no solutions.
A: Supose there is some $x\in\mathbb{R}$ such that $f'(x)=1/2$.
Since $f$ is continuously differentiable on $\mathbb{R}$ we have that $f'(x)=lim_{y\rightarrow x}\frac{f(x)-f(y)}{x-y}$
but $\frac{f(x)-f(y)}{x-y}$ is greater or equal than 1 or less or equal than -1 for every $x$ and $y$ in $\mathbb{R}$, by hypothesis. So this expression cannot converge to 1/2, so we have a contradiction! Hence, there is no such point $x$ in $\mathbb{R}$ such that $f'(x)=1/2$
