Intermediate value theorem: show there exists c ∈ [ 0 , 2 / 3 ] such that f ( c + 1 3 ) = f ( c )Let f : [ 0 , 1 ] → [ 0 , 1 ] be continuous and f ( 0 ) = f ( 1 )This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.Letting g ( x ) = f ( x + 1 3 ) − f ( x ) then this is only defined for the interval [ 0 , 2 3 ] b/c not sure what g ( 1 ) = f ( 4 3 ) − f ( 1 ) may be.Now, g ( 0 ) = f ( 1 3 ) − f ( 0 )and g ( 2 3 ) = f ( 1 ) − f ( 2 3 ) = f ( 0 ) − f ( 2 3 ).I can't (at least I don't see why I can) conclude this is equal to − g ( 0 ) which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.

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Intermediate value theorem: show there exists   c  ∈  [  0  ,  2      /    3  ] such that   f  (  c  +      1    3    )  =  f  (  c  )Let   f  :  [  0  ,  1  ]  →  [  0  ,  1  ] be continuous and   f  (  0  )  =  f  (  1  )This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.Letting   g  (  x  )  =  f  (  x  +      1    3    )  −  f  (  x  ) then this is only defined for the interval   [  0  ,      2    3    ] b/c not sure what   g  (  1  )  =  f  (      4    3    )  −  f  (  1  ) may be.Now,   g  (  0  )  =  f  (      1    3    )  −  f  (  0  )and   g  (      2    3    )  =  f  (  1  )  −  f  (      2    3    )  =  f  (  0  )  −  f  (      2    3    ).I can&#039;t (at least I don&#039;t see why I can) conclude this is equal to   −  g  (  0  ) which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.

kilinumad · Accepted Answer

HINT:You were on the right track by defining   g  (  x  )  ≡  f  (  x  +  1      /    3  )  −  f  (  x  ) for   x  ∈  [  0  ,  2      /    3  ].Now, note that  g  (  0  )  +  g  (  1      /    3  )  +  g  (  2      /    3  )  =  f  (  1  )  −  f  (  0  )  =  0Then either   g  (  0  )  =  g  (  1      /    3  )  =  g  (  2      /    3  )  =  0 or at least one of the three is positive and one of the three is negative. Now, use the IVT.

Intermediate value theorem: show there exists c in [0,2/3] such that f(c+1/3)=f(c)

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