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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How to prove this function has Darboux&#039;s property?  f  (  x  )  =      {                            cos          ⁡          (                      1            x                    )                                      if                           x              ≠              0                                                            0                                      if                           x              =              0                                              How do I prove this function has Darboux&#039;s property? I know it has it because it has antiderivatives, but how do I prove it otherwise, with intervals maybe?

Ordettyreomqu · Accepted Answer

Step 1Take,   a  ,  b  ∈      R   with   a  &amp;lt;  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  &amp;lt;  0 or   a  &amp;gt;  0, this is clear, by continuity. If   a  &amp;lt;  0  &amp;lt;  b, take some   n  ∈      N   such that       1          2      π      n        &amp;lt;  b. Step 2Then   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.

grenivkah3z · Accepted Answer

Step 1It&#039;s obvious that it has Darboux&#039;s property on any interval [a, b] with   0  &amp;lt;  a  &amp;lt;  b  ∨  a  &amp;lt;  b  &amp;lt;  0.If   a  ≤  0  &amp;lt;  b, then   f  (  a  )  ,  f  (  b  )  ∈  [  −  1  ,  1  ] anyways and there are some   0  &amp;lt;      x    1    &amp;lt;      x    2    &amp;lt;  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 2If   a  &amp;lt;  0  ≤  b, then   f  (  a  )  ,  f  (  b  )  ∈  [  −  1  ,  1  ] anyways and there are some   a  &amp;lt;      x    1    &amp;lt;      x    2    &amp;lt;  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  ).

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

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