Suppose that f is a continuous function on [0, 2] such that f(0) = f(2). Show that there is a real number ξ ∈ [1, 2] with f(ξ) = f(ξ − 1).
ξ ∈ [1, 2], ξ-1 ∈ [0, 1] Case 1: if f(0)=f(1) then ξ = 2, f(ξ) = f(0)= f(1) = f(2) therefore f(ξ)= f(ξ-1)=f(2)=f(1)
Case 2: if f(0)>f(1) then f(2) > f(1). By continuity there must exist some m where f(2) ≥ m ≥ f(1) and by IMV theorem there exists some ξ ∈ [1, 2] where f(ξ)=m. Now, ξ-1 ∈ [0, 1] and by continuity there must also exist the same m, f(0)≥ m ≥ f(1). Therefore by IMV theorem again there must exist at least 1 value where f(ξ − 1) = m and since m is common there must exist at least 1 value where f(ξ − 1) = f(ξ).
Case 3: f(1)>f(0). This is pretty much the same thing as case 2.
-This is a very difficult calculus question and I've never written a proof and I'm in Algebra II. Pls don't flame me if this is very stupid.
Your Case 2 proof, and therefore Case 3 proof too, looks dodgy. You correctly find $\xi\in[1,2]$ with $f(\xi) = m$. Also correctly, you note that $m$ is in the image of $[0,1]$ under $f$: there is some $\gamma\in[0,1]$ with $f(\gamma) = m$. You can even write $\gamma$ as $\xi'-1$ for some $\xi'\in[1,2]$. However, you're not entitled to conclude that $\xi' = \xi$.
A simpler approach: consider the function $g\colon[1,2]\rightarrow\mathbb{R}: \xi\mapsto f(\xi) - f(\xi-1)$. Then $g(2) = - g(1)$, so by continuity and IMV there is $\xi \in[1,2]$ such that $g(\xi) = 0$.