Let's define f as a continuous function with f : [ 0 ; 2 ] → R and f ( 0 ) = f ( 2 ).Now, I want to show that: ∃ x 0 ∈ [ 0 ; 1 ] : f ( x 0 ) = f ( x 0 + 1 )I tried to plot a few functions in order to construct a counterexample, but it seems that this statement really is true.Unfortunately, I don't think I'm entirely sure why this works, yet. My current guess is that it has something to do with the Intermediate Value Theorem, as f is continuous. In other words: If our function value 'goes up', it'll have to 'come back down' eventually (and vice versa), since it still needs to fulfill f ( 0 ) = f ( 2 ).Can someone help me prove this statement?

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Let&#039;s define   f as a continuous function with   f  :  [  0  ;  2  ]  →      R   and   f  (  0  )  =  f  (  2  ).Now, I want to show that:  ∃      x    0    ∈  [  0  ;  1  ]  :  f  (      x    0    )  =  f  (      x    0    +  1  )I tried to plot a few functions in order to construct a counterexample, but it seems that this statement really is true.Unfortunately, I don&#039;t think I&#039;m entirely sure why this works, yet. My current guess is that it has something to do with the Intermediate Value Theorem, as f is continuous. In other words: If our function value &#039;goes up&#039;, it&#039;ll have to &#039;come back down&#039; eventually (and vice versa), since it still needs to fulfill   f  (  0  )  =  f  (  2  ).Can someone help me prove this statement?

humbast2 · Accepted Answer

Consider the following function on [0,1],  g  (  x  )  =  f  (  x  +  1  )  −  f  (  x  )where we assume   f  (  0  )  =  f  (  2  ). We&#039;d like to show there is some       x    0    ∈  [  0  ,  1  ] such that   g  (      x    0    )  =  0. So if we can show   g  (  x  )  &amp;gt;  0 on some interval and   g  (  x  )  &amp;lt;  0 somewhere as well, since   g  (  x  ) is continuous it must pass 0. This is the context of the intermediate value theorem. So let&#039;s consider x=0 and 1, we see that  g  (  0  )  =  f  (  1  )  −  f  (  0  )    &amp;amp;    g  (  1  )  =  f  (  2  )  −  f  (  1  )  =  f  (  0  )  −  f  (  1  )  =  −  g  (  0  )Thus we see that   g  (  x  ) does indeed change sign. So by the intermediate value theorem we know that it must be zero somewhere.

Feinsn · Accepted Answer

Define  g  (  x  )  =  f  (  x  )  −  f  (  x  +  1  )for   x in [0,1] then you have that in 0 is positive but in 1 is negative and then by the MVT is done.

Let's define f as a continuous function with f : [ 0 ; 2 ] →<!

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