How to prove this function has Darboux's property? f ( x ) = { <mtabl

Agostarawz 2022-07-02 Answered
How to prove this function has Darboux's property?
f ( x ) = { cos ( 1 x ) if  x 0 0 if  x = 0
How do I prove this function has Darboux's property? I know it has it because it has antiderivatives, but how do I prove it otherwise, with intervals maybe?
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Answers (2)

Ordettyreomqu
Answered 2022-07-03 Author has 22 answers
Step 1
Take, a , b R with a < b. You want to prove that, if y lies between f(a) and f(b), then there is some c [ a , b ] such that f ( c ) = y. If b < 0 or a > 0, this is clear, by continuity. If a < 0 < b, take some n N such that 1 2 π n < b.
Step 2
Then f ( [ 1 2 π n + π , 1 2 π n ] ) = [ 1 , 1 ], and therefore there is come c [ 1 2 π n + π , 1 2 π n ] such that f ( c ) = y. The remaining cases are similar.
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grenivkah3z
Answered 2022-07-04 Author has 6 answers
Step 1
It's obvious that it has Darboux's property on any interval [a, b] with 0 < a < b a < b < 0.
If a 0 < b, then f ( a ) , f ( b ) [ 1 , 1 ] anyways and there are some 0 < x 1 < x 2 < b such that f ( x 1 ) = 1 and f ( x 2 ) = 1. By the previous remark, every value between -1 and 1 (so a fortiori every value between min { f ( a ) , f ( b ) } and max { f ( a ) , f ( b ) }) is attained in the interval [ x 1 , x 2 ] ( a , b ).
Step 2
If a < 0 b, then f ( a ) , f ( b ) [ 1 , 1 ] anyways and there are some a < x 1 < x 2 < 0 such that f ( x 1 ) = 1 and f ( x 2 ) = 1. For the same reason as the previous case, every value between -1 and 1 is attained in the interval [ x 1 , x 2 ] ( a , b ).
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