Does $\exists$ a differentiable function $f:\mathbb{R}\to \mathbb{R}, f(x) \neq x+c$ s.t. every interval $(a,b)$ contains a point $p$ with gradient 1? Does there exist a differentiable function $f:\mathbb{R} \to \mathbb{R}, f(x) \neq x+c$ such that every interval $(a,b)$ contains a point $p$ with gradient $1$?
I would guess no, but I've no idea how to prove it.
Now that I think about it, isn't this question similar to: Does there exist a non-constant differentiable function $f:\mathbb{R} \to \mathbb{R}$, such that every interval $(a,b)$ contains a point $p$ with gradient $0$? Which I may or may not have seen on the site somewhere - can't remember. I guess the disproof would be something like: $f(x)$ has gradient $0$ almost everywhere $\implies f(x)$ has unbounded variation somewhere (e.g. in some interval) $\implies$ f(x) is not everywhere differentiable. I'm not that familiar with bounded variation other than having skim-read this thread once, but bounded variation may in fact not be necessary to answer this question, I've no idea.
Maybe the mean value theorem is more relevant here.
 A: The answer is that yes, such a function exists.
The starting point is the Pompeiu Derivative..  To summarize Wikipedia, Pompeiu constructed an everywhere differentiable strictly increasing function $g:[0,1]\rightarrow \mathbb{R}$ whose derivative $g'(x)$ is $0$ on a dense subset of $[0,1]$.  Call this dense set $D$.
Now, let $h:\mathbb{R}\rightarrow (0,1)$ be your favorite diffeomorphism (e.g., you could pick $h(x) = \frac{1}{\pi}\arctan(x) + \frac{1}{2}$.)
Set $f(x) = g(h(x)) + x$.   I claim that $f$ fulfills your criterion.  By the chain rule, $f'(x) = g'(h(x))h'(x) + 1$.  For $x\in h^{-1}(D)$, $g'(h(x)) = 0$, so $f'(x) = 1$ for $x\in h^{-1}(D)$.  Note that since $h$ is a diffeomorphism, it is a homeomorphism, so $h^{-1}(D)\subseteq \mathbb{R}$ is dense.  Lastly, since $g$ is strictly increasing, $f(x) -x$ is strictly increasing, so is not constant.
A: (Sorry I don't have not enough reputation points to comment instead of answering...)
I don't have the answer but you should get a look at Darboux's theorem and Darboux's functions or similar functions. I think you can find a non linear function $f$ such that the set $\{ x |f^\prime(x)=1\}$ is dense in $\mathbb{R}$.
Note that you can change your problem by asking that $\{ x |f^\prime(x)=0\}$ is dense (consider $x \mapsto f(x)-x$). This could help together with the mean value Darboux's theorem
A: Is c a constant in (a,b) interval? If yes, you can try to prove it by contradiction, like using the function f(x) = x^2 in interval ]-1, 1[. I guess that's what you want.(Sorry about formatting, I'm ne here)
