# Suppose that $f(0)=f(2\pi)$. Show that there exists an x such that $f(x)=f(x+\pi)$.

I am supposed to show that there exists an $x$ in the interval $[0,\pi]$ such that $f(x)=f(x+\pi)$ by considering another function $g:[0,\pi] \to \mathbb{R}$ defined by $g(x)=f(x)-f(x+\pi)$.

Should I use a concrete example by considering a trig function?

• Sure sounds like a job for the intermediate value theorem Feb 3, 2016 at 23:28
• Concrete is often good. I would say not in this case. Feb 3, 2016 at 23:29
• What can be said about the values of $g(0)$ and $g(\pi)$? Feb 3, 2016 at 23:30
• What goes up must come down. Feb 3, 2016 at 23:40
• @George I think this is a very reasonable question to ask, even if it is an assignment question. OP has not asked for us to do the question entirely. They have shown us their thoughts and ideas (considering a trig function here is a very natural path to take in this question, even if it is not the actual way to go). Feb 4, 2016 at 0:38

As far as actually making the proof, try considering the intermediate value theorem. Note that $g(0) = f(0) - f(\pi)$ and $g(\pi) = f(\pi) - f(0) = - (f(0) - f(\pi))$. What can you conclude from this?
Consider $g(y)=f(y)-f(y+\pi)$ where $y\in [0,\pi]$. $g$ is continuous $g(0)=f(0)-f(\pi)=f(2\pi)-f(\pi)$ and $g(\pi)=f(\pi)-f(2\pi)=-g(0)$.
So $g(0)$ and $g(\pi)$ are of opposite sign and $g$ continuous. So by the intermediate value theorem
$$\exists x\in[0,\pi],\,g(x)=f(x)-f(x+\pi)=0$$