Exercise · ZW-EXTREME-VALUE-THEOREM-019 · ★★★ Advanced

Let \(f(x)=x\) if \(0\le x<1\) and \(f(1)=0\), defined on \([0,1]\). Which hypothesis of the theorem fails and what happens?

Choose one option.

Show answer
  • B. \(f\) is not continuous at \(1\); it does not attain its supremum \(1\)

Explanation

The interval is closed and bounded. Continuity fails at \(x=1\): \(\sup f=1\) is not attained, since \(f(x)<1\) always.

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Review the concept: Extreme value theorem

If \(f\) is continuous on a closed and bounded interval \([a,b]\), then \(f\) attains an absolute maximum and minimum on \([a,b]\).

\[f\ \text{continuous on }[a,b]\ \Rightarrow\ \exists\,m,M\in[a,b]:\ f(m)\le f(x)\le f(M)\]

More practice problems: extreme value theorem

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