Use the Intermediate Value Theorem to prove f : [ 0 , 1 ] → [ 0 , 1 ] continuous and C ∈ [ 0 , 1 ], there is some c ∈ [ 0 , 1 ] such that f ( c ) = C.Using a similar technique to the proof of the intermediate value theorem, I can easily prove that there is an f ( x ) = C, but I am having trouble proving that a f ( c ) = C.This is what I have:Since f is continuous on [0,1] there exists f ( a ) = 0 and f ( b ) = 1.Let g ( x ) = F ( x ) − C. We assume that f ( a ) &lt; C &lt; f ( b ) → 0 &lt; C &lt; 1when x = a, g ( a ) is negative when x = b, g ( b ) is positiveTherefore, g ( a ) &lt; 0 &lt; g ( b ), and since f is continuous on [0,1], so is g.Therefore there exists a g ( x ) = 0, and f ( x ) = C.How can I prove there is a f ( c ) = C ? Is this a rule for IVT?

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Use the Intermediate Value Theorem to prove   f  :  [  0  ,  1  ]  →  [  0  ,  1  ] continuous and   C  ∈  [  0  ,  1  ], there is some   c  ∈  [  0  ,  1  ] such that   f  (  c  )  =  C.Using a similar technique to the proof of the intermediate value theorem, I can easily prove that there is an   f  (  x  )  =  C, but I am having trouble proving that a   f  (  c  )  =  C.This is what I have:Since   f is continuous on [0,1] there exists   f  (  a  )  =  0 and   f  (  b  )  =  1.Let   g  (  x  )  =  F  (  x  )  −  C. We assume that   f  (  a  )  &amp;lt;  C  &amp;lt;  f  (  b  )   →   0  &amp;lt;  C  &amp;lt;  1when   x  =  a,   g  (  a  ) is negative when   x  =  b,   g  (  b  ) is positiveTherefore,   g  (  a  )  &amp;lt;  0  &amp;lt;  g  (  b  ), and since   f is continuous on [0,1], so is   g.Therefore there exists a   g  (  x  )  =  0, and   f  (  x  )  =  C.How can I prove there is a   f  (  c  )  =  C ? Is this a rule for IVT?

Ezequiel Davidson · Accepted Answer

The problem in your approach is that your already fix   C when you define   g, whereas we don&#039;t know a priori what it will be (and there is also a confusion between   c and   C).Consider   g  (  x  )  =  f  (  x  )  −  x;   g is continuous on [0,1], and   g  (  0  )  =  f  (  0  )  ≥  0, and   g  (  1  )  =  f  (  1  )  −  1  ≤  0 so   g take the value 0 at some   c  ∈  [  0  ,  1  ].

analianopolisca · Accepted Answer

You can prove using the intermediate value theorem that if   f  :  [  0  ,  1  ]  →  [  0  ,  1  ] is a continuous function then for every  λ  ∈      [          min              [        0        ,        1        ]              f    (    x    )    ,          max              [        0        ,        1        ]              f    (    x    )    ]  and only for these values of   λ, there exists   c  ∈  [  0  ,  1  ] with   f  (  c  )  =  λ.The problem is that you pick   C arbitrarily, and you can easily define continuous functions on [0,1] which do not take a certain value: constant functions, continuous functions   f  :  [  0  ,  1  ]  →  [  0  ,  C      /    2  ], etc.The flaw in your argument is that you assume that if   f is continuous then there exist   a  ,  b such that   f  (  a  )  =  0 and   f  (  b  )  =  1. You cannot assume such thing unless   f is onto

Use the Intermediate Value Theorem to prove f:[0,1]->[0,1] continuous and C in [0,1], there is some c in [0,1] such that f(c)=C.

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