Trying to take a different approach in proving if f : [ a , b ] → R is continuous and injective, then the maximum value of f must occur at a or b. trying to prove this by using the intermediate value theorem.Here is what I have so far:Let b ∈ R Assume to the contrary that the max value occurs at an interior point of the interval [ a , b ]. Let x ∘ equal this interior point.The first case would occur when f ( a ) &lt; f ( b ) then f ( a ) &lt; f ( b ) &lt; f ( x ∘ ) If we consider y ∘ ∈ ( f ( b ) , f ( x ∘ ) ) is there a way to show that y ∘ has at least 2 preimages by applying the intermediate value theorem?Of course the second case would be f ( b ) &lt; f ( a )...

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Trying to take a different approach in proving if   f  :  [  a  ,  b  ]  →      R   is continuous and injective, then the maximum value of   f must occur at   a or   b. trying to prove this by using the intermediate value theorem.Here is what I have so far:Let   b  ∈      R   Assume to the contrary that the max value occurs at an interior point of the interval   [  a  ,  b  ]. Let       x    ∘   equal this interior point.The first case would occur when   f  (  a  )  &amp;lt;  f  (  b  ) then   f  (  a  )  &amp;lt;  f  (  b  )  &amp;lt;  f  (      x    ∘    ) If we consider       y    ∘    ∈  (  f  (  b  )  ,  f  (      x    ∘    )  ) is there a way to show that       y    ∘   has at least 2 preimages by applying the intermediate value theorem?Of course the second case would be   f  (  b  )  &amp;lt;  f  (  a  )...

Sauppypefpg · Accepted Answer

Yes! You&#039;re very close. Notice first that you only really need to show this for the first case by symmetry, since the second case is virtually the same. Or instead, just say   c  =  max  (  f  (  a  )  ,  f  (  b  )  ) and deal with both at once.Now, take   z  ∈  (  c  ,  f  (      x    0    )  ). We have that   f  (  a  )  &amp;lt;  z  &amp;lt;  f  (      x    0    ), and   f  (  b  )  &amp;lt;  z  &amp;lt;  f  (      x    0    ), so what can we say about the preimage of   z?

Juan Leonard · Accepted Answer

Then of course       y    0    ∈  (  f  (  a  )  ,  f  (      x    0    )  ) as well! (Remember   f  (  a  )  &amp;lt;  f  (  b  )  &amp;lt;  f  (      x    0    ).)Thus there is   α between   a and       x    0   and another   β between       x    0   and   b such that   f  (  α  )  =  f  (  β  )  =      y    0  , so   f is not injective.You were really close to the proof.

Trying to take a different approach in proving if f:[a,b]->R is continuous and injective, then the maximum value of f must occur at a or b. trying to prove this by using the intermediate value theorem.

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