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

Let $$f,g:[0,1]\to \mathbb{R}$$ be two functions of class $$C^1$$ such that $$f'(x)>0$$ and $$g'(x)>0$$ for all $$x\in [0,1]$$. Assume that $$f(0) = g(0) \qquad \text{and} \qquad f(1) = g(1).$$ Show that there exist $$x,y\in[0,1]$$ such that $$f(x) = g(y) \qquad \text{and} \qquad f'(x) = g'(y).$$

I made the following attempts (non of them worked): First, define a function $$h:[0,1]\times [0,1]\to \mathbb{R}$$ given by $$h(x,y) = [f(x)-g(y)]^2 + [f'(x)-g'(y)]^2$$ and try to show that $$h(x,y) = 0$$ for some $$(x,y)$$. Then I defined $$k:[0,1]\times [0,1] \to \mathbb{R}$$ by $$k(x,y) = \frac{1}{2}[f(x)-f(y)]^2$$ and tried to show that at some point $$(x,y)$$ the divergence of $$k$$ is zero, because in this case we have that $$0 = \frac{\partial k}{\partial x}(x,y) + \frac{\partial k}{\partial y}(x,y) = (f(x)-g(y))f'(x)-(f(x)-f(y))g'(y) = (f(x)-f(y))(f'(x)-g'(y))$$ but it didn't work either. May be I'm not following the right way to solve this exercise.

• Shouldn't it rather be $x,y\in [0,1]$? Commented Nov 8, 2020 at 20:37
• Sorry people, I just assumed that "two points" meant that the points had to be distinct. In fact, it does; but the correct edit would have been to remove the words "two points". Also, $x\le y$ is irrelevant, so the correct correct edit would have been to replace "two points $x\le y$" with "$x$ and $y$". (And I see now that @popoolmica got there before me.) Commented Nov 8, 2020 at 20:38
• Sorry to all of you. May be I made a bad translation (from spanish) of the original exercise. Commented Nov 8, 2020 at 20:43
• It seems to me that in some places [0,1] should be (0,1) (as in open interval), right? You can't really talk about the derivative at the endpoint of the interval of definition, and anyway Rolle's Theorem requires differentiability only in the interior. Commented Nov 9, 2020 at 9:55
• @AriBrodsky: It is well-defined what it means for a function to be differentiable on a closed interval, including the endpoints. But you are right that the conditions can be relaxed to only require that $f, g$ are continuous in the closed and differentiable in the open interval. The given answers also show that it is possible to find $x, y$ in the open interval $(0, 1)$. Commented Nov 9, 2020 at 11:02

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.