Let $f : R \to R$ be a continuously differentiable function such that $|f(x)-f(y)| \geq |x-y$| for all x,y $\in R$. Let $f : R \to R$ be a continuously differentiable function such that $|f(x)-f(y)| \geq |x-y$| for all x,y $\in R$. Then i have to find number of solutions of $f'(x)=   \frac {1}{2} $
Attempt
I thought of using mean value theorem.i write as
$\frac{|f(x)-f(y)|}{|x-y|} = |f'(c)|$, where c lies between x and y
Now to satisfy with question i need to have $f'(c) \geq 1$. So how can it be $\frac{1}{2}$. Is this right way to do this?
Thanks

Pose $x = t + h$ and $y = t$. You get $\frac{|f(t+h) - f(t)|}{|h|} \geq 1$ for every $t,h$ real numbers. Take the limit $h \to 0$, you get $f'(t) \geq 1$ for every $t$ real. So there are no solutions for the requested equation.
