Let I be a real interval. Let a,b in I such that (a..b) is an open interval. Let f:I->R be a real function which is continuous on (a..b). Let k in R lie between f(a) and f(b). That is, either f(a)<k<f(b) or f(b)<k<f(a). Then EEc in (a..b) such that f(c)=k.

Seettiffrourfk6 2022-10-11 Answered
Found it:
Let I be a real interval. Let a , b I such that ( a . . b ) is an open interval. Let f : I R be a real function which is continuous on ( a . . b ). Let k R lie between f ( a ) and f ( b ). That is, either f ( a ) < k < f ( b ) or f ( b ) < k < f ( a ). Then c ( a . . b ) such that f ( c ) = k.

Shouldn't this suppose continuity on the closed interval [ a , b ]? Otherwise consider f : [ 1 , 1 ] R such that
f ( x ) = { arcsin ( x ) -1<x<1 23 if x=1 23 if x= -1
where a = 1 and b = 1.
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Answers (1)

plomet6a
Answered 2022-10-12 Author has 20 answers
Yes, I agree. The given statement is incorrect and your counterexample shows it.

An even simpler example might just be
f ( x ) = { 0 , x = 1 2 , 1 < x < 1 1 , x = 1
with k = 1 2 .
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