Is there exists a $y$ in the interval $(0,1)$ such that $f(y)=f(y+1) ?$

Question

A function $f(x)$ is continuous in the interval $[0,2].$ It is known that $f(0)=f(2)=−1$ and $f(1)=1.$ Which one of the following statements must be true$?$

Options are $:$

1. There exists a $y$ in the interval $(0,1)$ such that $f(y)=f(y+1)$
2. For every $y$ in the interval $(0,1),f(y) = f(2−y)$
3. The maximum value of the function in the interval $(0,2)$ is $1$
4. There exists a $y$ in the interval $(0,1)$ such that $f(y) = −f(2−y)$

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I try to explain $:$

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Since the function is continuous, there must be point between $0$ and $1$ where it becomes $0$ and there must also be a point between $1$ and $2$ where it becomes $0.$ .

Option $(1)$ should be true at $y=0$.

Can you explain in formal way ?

Answer 1

Let a new function $ g(y) = f(y+1)-f(y) $ defined in $y$ in $ [0,1] $. $ g(0) = f(1)-f(0) $ and $ g(1) = f(2) -f(1) = f(0)- f(1) = -g(0) $. As the function is continuous, and it changes sign. $ g(x) =0 $ for some $ x $ in $ (0,1) $. Option $ 1 $ is true.