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.

rivasguss9

rivasguss9

Open question

2022-08-13

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 ) < C < f ( b ) 0 < C < 1
when x = a, g ( a ) is negative when x = b, g ( b ) is positive
Therefore, g ( a ) < 0 < 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?

Answer & Explanation

Ezequiel Davidson

Ezequiel Davidson

Beginner2022-08-14Added 11 answers

The problem in your approach is that your already fix C when you define g, whereas we don'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

analianopolisca

Beginner2022-08-15Added 2 answers

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

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