I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed:Suppose f : [ − 1 , 1 ] → R is continuous and satisfies f ( − 1 ) = f ( 1 ). Prove that there exists y ∈ [ 0 , 1 ] such that f ( y ) = f ( y − 1 ).So far, I have consideblack a new function g ( x ) = f ( x ) − f ( x − 1 ), x ∈ [ 0 , 1 ], but am stuck after this.

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I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed:Suppose   f  :  [  −  1  ,  1  ]  →      R   is continuous and satisfies   f  (  −  1  )  =  f  (  1  ). Prove that there exists   y  ∈  [  0  ,  1  ] such that   f  (  y  )  =  f  (  y  −  1  ).So far, I have consideblack a new function   g  (  x  )  =  f  (  x  )  −  f  (  x  −  1  ),   x  ∈  [  0  ,  1  ], but am stuck after this.

renegadeo41u · Accepted Answer

Using the fact that   f  (  1  )  =  f  (  −  1  ), we have  g  (  0  )  =  f  (  0  )  −  f  (  −  1  )  =  f  (  0  )  −  f  (  1  )  g  (  1  )  =  f  (  1  )  −  f  (  0  )  =  −  [  f  (  0  )  −  f  (  1  )  ]So,   g  (  0  ) and   g  (  1  ) have opposite signs. By continuity of   g, it must have a root in [0,1], so there exists   y  ∈  [  0  ,  1  ] such that   g  (  y  )  =  0, i.e.   f  (  y  )  =  f  (  y  −  1  ).

I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed: Suppose f:[−1,1]->R is continuous and satisfies f(−1)=f(1). Prove that there exists y in [0,1] such that f(y)=f(y−1).

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