# Let f:[0,2]->R be a continuous function such that f(0)=f(2). Use the intermediate value theorem to prove that there exist numbers x,y in[0,2] such that f(x)=f(y) and |x−y|=1. Hint: Introduce the auxiliary function g:[0,1]->R defined by g(x)=f(x+1)−f(x).

Let $f:\left[0,2\right]\to \mathbb{R}$ be a continuous function such that $f\left(0\right)=f\left(2\right)$. Use the intermediate value theorem to prove that there exist numbers $x,y\in \left[0,2\right]$ such that $f\left(x\right)=f\left(y\right)$ and $|x-y|=1$.
Hint: Introduce the auxiliary function $g:\left[0,1\right]\to \mathbb{R}$ defined by $g\left(x\right)=f\left(x+1\right)-f\left(x\right)$.
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The given function $\phantom{\rule{thickmathspace}{0ex}}g\phantom{\rule{thickmathspace}{0ex}}$ is continous in $\phantom{\rule{thickmathspace}{0ex}}\left[0,1\right]\phantom{\rule{thickmathspace}{0ex}}$ and
$\left\{\begin{array}{l}g\left(0\right)=f\left(1\right)-f\left(0\right)\\ \\ g\left(1\right)=f\left(2\right)-f\left(1\right)\end{array}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}g\left(0\right)g\left(1\right)<0\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{, unless}\phantom{\rule{thickmathspace}{0ex}}f\left(1\right)=f\left(0\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{(why?)}$
and so by the IVM there exists $\phantom{\rule{thickmathspace}{0ex}}c\in \left(0,1\right)\phantom{\rule{thickmathspace}{0ex}}$ s.t. $\phantom{\rule{thickmathspace}{0ex}}g\left(c\right)=0\phantom{\rule{thickmathspace}{0ex}}\dots$ end the exercise now,