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I have attached a question and answer below. The answer does not really make sense to me especially the underlined parts. I must be missing something simple I would appreciate if someone could help me clear things out. enter image description here

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You’re not missing anything: the answer is nonsense.

Added: I have to take that back: Ewan Delanoy has found what was probably the intended argument. It strikes me as doing things the hard way, but it does work.

After the points $x$ and $y$ are chosen with $x<y$ and $f(x)<f(y)$, I would continue by looking at the interval $[x,b]$. We know that $f(x)<f(y)>f(b)$, so the maximum of $f$ on $[x,b]$ is not at $x$ or $b$, and therefore $f$ has a local maximum at some $z\in(x,b)$, contradicting the hypothesis.

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  • $\begingroup$ Thanks Brian I find both the answers I have received very useful. $\endgroup$ – Heisenberg Nov 10 '13 at 15:50
  • $\begingroup$ @Rajinda: You’re welcome. $\endgroup$ – Brian M. Scott Nov 10 '13 at 15:50
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What happens is this :

  • ${\min}_{[a,y]}f=f(y)$, so $f(t) \geq f(y)$ for all $t\in[x,y]$.

  • ${\max}_{[x,y]}f=f(y)$, so $f(t) \leq f(y)$ for all $t\in[x,y]$.

So we have $f(t)=f(y)$ for all $t\in [x,y]$, so $f$ is constant on $[x,y]$.

Then any $z\in (x,y)$ is both a local minimum/maximum.

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  • $\begingroup$ Thanks a lot Ewan for the explanation $\endgroup$ – Heisenberg Nov 10 '13 at 15:50
  • $\begingroup$ You’re welcome. $\endgroup$ – Ewan Delanoy Nov 10 '13 at 15:51

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