The Intermediate Value Theorem states that, for a continuous function f : [ a , b ] → R , if f ( a ) &lt; d &lt; f ( b ), then there exists a c ∈ ( a , b ) such that f ( c ) = d .I wonder if I change the hypothesis of f ( a ) &lt; d &lt; f ( b ) to f ( a ) &gt; d &gt; f ( b ), the result still holds. I believe so, since f assumes a fixed point f ( x ) = x in [ a , b ], so we would have c = d, although I'm not completely sure.I need this result in order to prove the set X = { x ∈ [ a , b ] s . t . f | [ a , x ] is bounded }, with a continuous f, is not empty.

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The Intermediate Value Theorem states that, for a continuous function   f  :  [  a  ,  b  ]  →      R  , if   f  (  a  )  &amp;lt;  d  &amp;lt;  f  (  b  ), then there exists a   c  ∈  (  a  ,  b  ) such that   f  (  c  )  =  d  .I wonder if I change the hypothesis of   f  (  a  )  &amp;lt;  d  &amp;lt;  f  (  b  ) to   f  (  a  )  &amp;gt;  d  &amp;gt;  f  (  b  ), the result still holds. I believe so, since   f assumes a fixed point   f  (  x  )  =  x in   [  a  ,  b  ], so we would have   c  =  d, although I&#039;m not completely sure.I need this result in order to prove the set   X  =  {  x  ∈  [  a  ,  b  ]    s  .  t  .    f      |    [  a  ,  x  ]        is bounded    }, with a continuous   f, is not empty.

Aryanna Caldwell · Accepted Answer

I don&#039;t know how you&#039;re getting that   f has a fixed point, or how that would imply that   c  =  d. You seem to be confusing inputs and outputs of   f: we know nothing at all about how   f  (  a  ) and   f  (  b  ) are related to   a and   b.However, you can instead just apply the Intermediate Value Theorem to the function   g  (  x  )  =  −  f  (  x  ), which satisfies   g  (  a  )  &amp;lt;  −  d  &amp;lt;  g  (  b  ), to get some   c such that   g  (  c  )  =  −  d, so   f  (  c  )  =  d.(It is not clear to me what your question has to do with proving that the set   X is nonempty, though.)

The Intermediate Value Theorem states that, for a continuous function f : [ a , b

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