Pair of functions with same values and same derivatives at distinct points

Set $a = f(0) = g(0)$ and $b = f(1) = g(1)$. $f$ and $g$ are strictly increasing from $[0, 1]$ to $[a, b]$ and therefore invertible. If we define the function $$ h: [a, b]\to \Bbb R, \, h(t) = f^{-1}(t) - g^{-1}(t) $$ then $h(a) = h(b) = 0$ and Rolle's theorem shows that for some $c \in (a, b)$ $$ 0 = h'(c) = \frac{1}{f'(f^{-1}(c))} - \frac{1}{g'(g^{-1}(c))} \\ \implies f'(f^{-1}(c)) = g'(g^{-1}(c)) \, . $$ We can now set $$ x = f^{-1}(c) \, , \, y = g^{-1}(c) \, . $$ $x$ and $y$ are both in the open interval $(0, 1)$ and have the desired properties $$ f(x) = g(y) \, , \, f'(x) = g'(y) \, . $$

Illustration: The graphs of $f$ and $g$ are green. $c$ is the argument on the $y$-axis where the inverse functions have the same slope. $x$ and $y$ are the pre-images of $c$ under $f$ and $g$.

(Created with GeoGebra.)

Remarks:

• If $f$ and $g$ are not identical functions then $h$ is not identically zero and we can choose $c\in (a, b)$ as a point where $h$ attains its nonzero minimum or maximum, so that $x - y = h(c) \ne 0$.

  This shows that (unless $f=g$) a pair $x, y$ of distinct points in $(0, 1)$ with those properties exists.

• It suffices to require that the functions are continuous on $[0, 1]$, and differentiable on $(0, 1)$ with positive derivative. The continuity of the derivative, or differentiability at the endpoints is not needed. (This was noticed by Ari Brodsky in a comment.)

Define the paths $\gamma_f : [0,1] \rightarrow \mathbb{R}^2$ and $\gamma_g : [0,1] \rightarrow \mathbb{R}^2$ by $$\forall x \in [0,1], \quad \gamma_f(x)=\left(f(x), \frac{1}{f'(x)}\right) \quad \quad \text{and} \quad \quad \gamma_g(x)=\left(g(x),\frac{1}{g'(x)}\right)$$

Because $f$ and $g$ are strictly increasing, then these two paths can be seen as the graph of two functions $F$ and $G$, defined on the interval $[a,b]$, where $a=f(0)=g(0)$ and $b=f(1)=g(1)$.

Moreover, using substitutions $t=f(u)$ and $y=g(u)$, one has $$\int_a^b F(t)dt = \int_0^1 \frac{f'(u)}{f'(u)} \mathrm{du} = 1 = \int_0^1 \frac{g'(u)}{g'(u)} \mathrm{du} = \int_a^b G(y)\mathrm{dy}$$

So the graphs of $F$ and $G$ intersect, which means that the paths $\gamma_f$ and $\gamma_g$ intersect, and you are done.