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Let $f$ be continuous on $[a,b]$ where $a<b$ such that $f(x)\ne0$ for all $x\in[a,b]$. Prove that there is a $c>0$ where either $f(x)>c$ for all $x\in[a,b]$ or $f(x)<-c$ for all $x\in[a,b]$.

Is this about local/absolute minimum and maximum? I do not know how to approach this question and any guidance to at least start the approach would be very much appreciated. Thank you!

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  • $\begingroup$ Do you know the properties of a function continous over an interval? $\endgroup$ – nls Sep 18 '16 at 9:45
  • $\begingroup$ 1) $f$ is bounded 2) $f$ has a minimum and maximum on $[a,b]$ $\endgroup$ – nls Sep 18 '16 at 9:46
  • $\begingroup$ @Ra You may want to google "Weierstrass Theorem on continuous functions" ... $\endgroup$ – DonAntonio Sep 18 '16 at 9:57
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Since $f$ is continuous, by the extreme value theorem there exists a lower bound $m$ and an upper bound $M$ on $[a,b]$. Let $q,Q\in[a,b]$ be such that $f(q)=m$ and $f(Q)=M$.

We have that both $m$ and $M$ are non-zero from the definition of $f$, so they are either positive or negative each. Suppose they are of opposite signs, then by the intermediate value theorem there exists some $r$ between $q$ and $Q$ with $f(r)=0$, but the definition of $f$ precludes this.

Hence $m$ and $M$ have the same signs. If they are both positive, we choose $c$ such that $0<c<m$; if they are both negative, we choose $c$ such that $M<-c<0$. Then $c$ or $-c$ is a lower or upper bound on $f$, respectively. This completes the proof.

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