Let f be continuous on [ a , b ], where a &lt; b, such that f ( x ) ≠ 0 for all x ∈ [ a , b ]. Prove that there is a c &gt; 0, either f ( x ) &gt; c or f ( x ) &lt; − c for all x ∈ [ a , b ].What I did,Proving by contradiction, Let f ( x ) &gt; c for some x 1 ∈ [ a , b ] and f ( x ) &lt; − c for some x 2 ∈ [ a , b ]. By the Intermediate Value Theorem, there exists exists a number z ∈ ( a , b ) such that f ( z ) = 0, however this contradicts the fact that f ( x ) ≠ 0 for all x ∈ [ a , b ]. Hence f ( x ) &gt; c or f ( x ) &lt; − c for all x ∈ [ a , b ].Hi, was wondering whether my proof is sound. Do help me verify! Thanks!

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Let   f be continuous on   [  a  ,  b  ], where   a  &amp;lt;  b, such that   f  (  x  )  ≠  0 for all   x  ∈  [  a  ,  b  ]. Prove that there is a   c  &amp;gt;  0, either   f  (  x  )  &amp;gt;  c or   f  (  x  )  &amp;lt;  −  c for all   x  ∈  [  a  ,  b  ].What I did,Proving by contradiction, Let   f  (  x  )  &amp;gt;  c for some       x          1        ∈  [  a  ,  b  ] and   f  (  x  )  &amp;lt;  −  c for some       x          2        ∈  [  a  ,  b  ]. By the Intermediate Value Theorem, there exists exists a number   z  ∈  (  a  ,  b  ) such that   f  (  z  )  =  0, however this contradicts the fact that   f  (  x  )  ≠  0 for all   x  ∈  [  a  ,  b  ]. Hence   f  (  x  )  &amp;gt;  c or   f  (  x  )  &amp;lt;  −  c for all   x  ∈  [  a  ,  b  ].Hi, was wondering whether my proof is sound. Do help me verify! Thanks!

Jewel Christensen · Accepted Answer

You got the negation of the statment wrong. The negation of &quot;There exists a   c  &amp;gt;  0 such that for all x   f  (  x  )  &amp;gt;  c or for all x   f  (  x  )  &amp;lt;  −  c&quot; is actually &quot;for each   c  &amp;gt;  0 there is an   x such that   f  (  x  )  &amp;lt;  c and there is an x such that   f  (  x  )  &amp;gt;  −  c&quot;. That is the statement from which you should derive a contradiction.Note: the point of the question is to use the fact that you are on a closed interval. On an open interval, for instance (0,1), the statement does not hold (just look at   f  (  x  )  =  1  −  x).

Let f be continuous on [ a , b ] , where a < b , such that

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