Suppose that f : [ 0 , 1 ] → R is a continuous function on the interval [ 0 , 1 ], and that f ( 0 ) = f ( 1 ). Use the Intermediate Value Theorem to show that there exists c ∈ [ 0 , 1 / 2 ] such that f ( c ) = f ( c + 1 / 2 ).I understand the Intermediate Value Theorem and I know how it is used to prove an equation has a root within an interval. I just don't understand how to approach this question

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Suppose that   f  :  [  0  ,  1  ]  →      R   is a continuous function on the interval   [  0  ,  1  ], and that   f  (  0  )  =  f  (  1  ). Use the Intermediate Value Theorem to show that there exists   c  ∈  [  0  ,  1      /    2  ] such that   f  (  c  )  =  f  (  c  +  1      /    2  ).I understand the Intermediate Value Theorem and I know how it is used to prove an equation has a root within an interval. I just don&#039;t understand how to approach this question

Zackery Harvey · Accepted Answer

We want   f  (  c  )  =  f  (  c  +      1    2    ), i.e.,   f  (  c  )  −  f  (  c  +      1    2    )  =  0. This already looks more like proving a function has a root. In fact, if we define a new function   g  (  c  )  =  f  (  c  )  −  f  (  c  +      1    2    ), then our job is exactly to prove   g has a root in   [  0  ,      1    2    ].We have to verify that   g is continuous in order to use the IVT, but once you do this, it becomes a much more standard IVT question, which you indicated you are familiar with.

Crystal Wheeler · Accepted Answer

Define   g  (  x  )  =  f  (  x  )  −  f  (  x  +  1      /    2  ). If   f  (  0  )  =  f  (  1      /    2  ), we&#039;re done; otherwise, WLOG   f  (  1      /    2  )  &amp;lt;  f  (  0  ). Thus   g  (  0  )  &amp;gt;  0 and   g  (  1      /    2  )  =  f  (  1      /    2  )  −  f  (  1  )  =  f  (  1      /    2  )  −  f  (  0  )  &amp;lt;  0; in particular, since   g is continuous (as a sum of continuous functions), we have that there exists a   c  ∈  [  0  ,  1      /    2  ] such that   g  (  c  )  =  0. Equivalently,   f  (  c  )  =  f  (  c  +  1      /    2  ).

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