Theorem: Let f be continuous on [a,b] and assume f(a)!=f(b). Then for every lambda such that f(a)<lambda<f(b), there exists a c in (a,b) such that f(c)=lambda. Question: Suppose that f:[0,1]->[0,2] is continuous. Use the Intermediate Value Theorem to prove that their exists c in [0,1] such that:

Leypoldon 2022-08-08 Answered
Theorem: Let f be continuous on [ a , b ] and assume f ( a ) f ( b ). Then for every λ such that f ( a ) < λ < f ( b ), there exists a c ( a , b ) such that f ( c ) = λ.

Question:
Suppose that f : [ 0 , 1 ] [ 0 , 2 ] is continuous. Use the Intermediate Value Theorem to prove that their exists c [ 0 , 1 ] such that:
f ( c ) = 2 c 2

Attempt:
I know that when we have the condition were f : [ a , b ] [ a , b ], the method to prove that c exits, is the same method you would use to prove the fixed point theorem.

Unfortunately I don't have an example in my notes when we have f : [ a , b ] [ a , y ]. How would I use the IVT to answer the original question?
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Answers (1)

pokajalaq1
Answered 2022-08-09 Author has 18 answers
Hint:
Consider the function g ( x ) = f ( x ) 2 x 2 .
Some details:
g ( 0 ) = f ( 0 ) 0 since the range of f is contained in [0,2]. Similarly, g ( 1 ) = f ( 1 ) 2 0. Furthermore, g is continuous since f is. Now, either g ( 0 ) or g ( 1 ) == 0, and there's nothing to prove. Or g ( 0 ) > 0, g ( 1 ) < 0. The Intermediate value theorem assures there exists c ( 0 , 1 ) such that g ( c ) = 0.
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