Is there a function whose graph intersects every tangent line at exactly 2 points? Is there some $f:\mathbb{R}\to\mathbb{R}$ differentiable at every point such that $\forall x$, the tangent to $f$ at $(x,f(x))$ intersects the graph of $f$ at $2$ points (counting (x,f(x)))?
This problem is delicate, as shown by the example $f(x)=x^3$, where the condition only fails at $x=0$.
Remark: if $f$ is $C^1$, then the function $g:\mathbb{R}\to\mathbb{R}$ which sends $x$ to the only point $y$ such that $(y,f(y))$ is in the tangent line to $x$ seems to be continuous at almost every point. I could not use this effectively though, in the non continuous case maybe you could get some weak version of this using the intermediate value property of the derivative?
 A: No $f$ as in the question exists.
Suppose $f$ is as in the statement of the question.
We can suppose that $f(0)=f'(0)=f(1)=0$ and $f>0$ in $(0,1)$ composing $f$ with affine functions or adding affine functions to it. $(*)$
We can then take $a\in(0,1)$ such that $f(a)$ is maximal and let $b\neq a$ with $f(b)=f(a)$. Then $b$ cannot be inside the interval $(0,1)$: in that case, the tangent at the point $c$ of the interval between $a$ and $b$ such that $f(c)$ is minimal cuts the graph of $f$ in at least $3$ points. There are two options for $b$.
If $b>1$, we can find a contradiction: in this case $f(1)=0$ is a local minimum of $f$. Now using the fact that for any $\varepsilon$ there is some $x\in(0,\varepsilon)$ with $f'(x)\in(0,\varepsilon)$ (you can prove it using that $f(0)=f'(0)=0$ and $f(x)>0$ in $(0,1)$), we can find a small $x>0$ such that its tangent line intersects the graph of $f$ at two points near $(1,0)$.
So $b<0$ and $f$ does not have a local minimum at $1$. The graph of $f$ would look something like this:

We have two cases:
Case 1: $f'$ is continuous.
Consider the point $c\in[0,a]$ such that $f'(c)$ is the highest possible, you can check that the tangent to $f$ at $c$, $l_c$, only intersects the graph at $(c,f(c))$. Indeed, if $(d,f(d))$ was another point in $l_c$, we can consider two cases.

*

*$d\in(0,a)$. Then there has to be some number $e$ between $c$ and $d$ with $f'(e)>f'(c)$, a contradiction.


*$d>a$ cannot happen either because the tangent line to $c$ passes above $(a,f(a))$ and $f(x)<f(a)\forall x>a$. A similar argument works for $d<0$.
Case 2: $f'$ is not continuous.
We can suppose that $f'$ is not continuous at $0$ in this case, using a point with discontinuous derivative in $(*)$.
As $(1,0)$ is not a local minimum, there exists some $\varepsilon_1>0$ such that $(-\varepsilon_1,\varepsilon_1)\subseteq f([a,1.1])$ $(**)$.
As $f'$ is not continuous at $0$ and using the intermediate value property of the derivative, for small enough $\varepsilon>0$ there is a sequence $x_n\to 0$ with $\varepsilon<|f'(x_n)|<2\varepsilon$. Let $l_n$ be the tangent line at $x_n$. Then, if we pick $\varepsilon$ small enough, for big enough $n$ the lines $l_n$:

*

*Pass below the points $(b,f(b))$ and $(a,f(a))$.


*Intersect the graph of $f$ at some point of the rectangle $[a,1.1]\times(-\varepsilon_1,\varepsilon_1)$, due to $(**)$
We can consider two cases. If $f'(x_n)<0$ for infinitely many $n$, then take some $x\in (-0.1,0)$ with $|\frac{f(x)}{x}|<\varepsilon$. Then for big enough $n$, $l_n$ intersects the graph of $f$ at some point between $b$ and $x$, which is a contradiction. If $f'(x_n)>0$ for infinitely many $n$, take $x\in (0,a)$ with $\frac{f(x)}{x}<\varepsilon$ and for big enough $n$, $l_n$ intersects the graph at some point between $x$ and $a$, so we have a contradiction again.
A: While I was working the problem (and it was a pretty interesting problem), a complete answer appeared, but I think I'm still going to post all the results that I got so far. Not all of them are necessary for the final solution, but still interesting nonetheless.
Assume, there exists a differentiable function $f$ as described. Let $g\colon\mathbb{R}\rightarrow\mathbb{R}$ be the fixed point free function as described, that assigns a value $x\in\mathbb{R}$ the abscissa of the unique other point, where the tangent to $f$ at $(x,f(x))$ intersects the graph of $f$. Therefore $g(x)$ is the unique value, so that:
\begin{equation}
f(x)+f'(x)(g(x)-x)=f(g(x))
\Leftrightarrow
f'(x)=\frac{f(g(x))-f(x)}{g(x)-x}.
\end{equation}
Conclusion: If $g$ is continuous in $x$, then $f'$ is continuous in $x$.
Proposition: Changing $f(x)$ to $f(x)+\lambda x+\mu$ (with the same sought-after property) doesn't change $g(x)$.
Corollary: The conditions $g(g(x))=x$ and $f'(g(x))=f'(x)$ are equivalent.
Proof: $\Rightarrow$: Take the upper equation for $x$ and $g(x)$, combine them through $f(g(x))$ and use that $g$ doesn't have a fixed point to shorten $g(x)-x$. $\Leftarrow$: Directly use the upper equation and expand the fraction by $-1$. $\square$
Lemma: If $g$ is not continuous in $x$, then either the equivalent conditions of the upper corollary hold or $f'$ is not continuous in $x$.
Notice, that when $f$ is continuously differentiable, the first case always has to hold and that the backwards direction from the second case is the contraposition of the upper conclusion.
Proof: We prove the contraposition. Assume $f'(g(x))\neq f'(x)$ and $f'$ is continuous in $x$. Define $\widetilde{f}(y):=f(y)-f'(x)(y-x)-f(x)$ with $\widetilde{f}(x)=0$ and $\widetilde{f}'(x)=0$ as well as $\widetilde{f}(g(x))=0$ using the upper equation and proposition. (Particularly $\widetilde{f}$ doesn't have any other roots than $x$ and $g(x)$.) $\widetilde{f}'$ is continuous in $x$ and the assumption translates to $\widetilde{f}'(g(x))\neq 0$, so for every neighborhood $U$ of $g(x)$, $\widetilde{f}(U)$ is a neighborhood of $\widetilde{f}(g(x))$. Take a sequence $x_n\xrightarrow{n\rightarrow\infty}x$, then $\widetilde{f}(x_n)\xrightarrow{n\rightarrow\infty}\widetilde{f}(x)=0$ and $\widetilde{f}'(x_n)\xrightarrow{n\rightarrow\infty}\widetilde{f}'(x)=0$. The tangents $t_n(y)$ through $(x_n,\widetilde{f}(x_n))$ are given by $t_n(y)=\widetilde{f}(x_n)+\widetilde{f}'(x_n)(y-x_n)$. We have $t_n(g(x))\xrightarrow{n\rightarrow\infty}0$, so for all $n$ above a sufficient high boundary, $t_n(g(x))\in\widetilde{f}(U)$ and therefore $g(x_n)\in U$ (Notice, that a higher boundary may be necessary.), which implies $g(x_n)\xrightarrow{n\rightarrow\infty}g(x)$. $\square$
If $g$ is not continuous in $x$ and the first case of this lemma holds, then $g(x)-x$ changes signs between $x$ and $g(x)$, so there has to be another point where $g$ is not continuous between them, otherwise $\widetilde{g}$ would have a root or equivalently $g$ a fixed point due to the intermediate value theorem.
Assume $x<g(x)$ w.l.o.g.  Because of the mean value theorem, there exists a $\xi\in(x,g(x))$ (which especially means $\widetilde{f}(\xi)\neq 0$) with $f'(\xi)=f'(x)\Leftrightarrow\widetilde{f}'(\xi)=0$ using the upper equation. Assume $\widetilde{f}$ is only positive in $(x,g(x))$ w.l.o.g. Let $\xi\in(x,g(x))$ be the value of the global maximum in $(x,g(x))$. It is unique since for two different maxima $\xi_1,\xi_2\in(x,g(x))$ with $\xi_1<\xi_2$ w.l.o.g., there would be a $\zeta\in(\xi_1,\xi_2)$ with $\widetilde{f}'(\zeta)=0$ and $\widetilde{f}(\zeta)<\widetilde{f}(\xi_1)=\widetilde{f}(\xi_2)$. Due to the intermediate value theorem, the horizontal tangent through $(\zeta,\widetilde{f}(\zeta))$ would intersect $\widetilde{f}$ also somewhere in $(x,\xi_1)$ and $(\xi_1,g(x))$. This also implies $g(\xi)\notin[x,g(x)]$. We are left with two cases for which we have to derive a contradiction:
Case 1: Assume $g(\xi)\in(-\infty,x)$, then $f>0$ in $(-\infty,0)$ and $f<f(\xi)$ in $(\xi,\infty)$.
Case 2: Assume $g(\xi)\in(g(x),\infty)$, then $f>0$ in $(g(x),\infty)$ and $f<f(\xi)$ in $(-\infty,x)$.
The argumentation for both cases is analogous and done in the other answer.

