Exercise · ZW-EXTREME-VALUE-THEOREM-013 · ★☆☆ Basic
Why does the theorem not apply to \(f(x)=1/x\) on \((0,1]\), even though \(f\) is continuous there?
Choose one option.
Show answer
- B. Because \((0,1]\) is not closed: it lacks the endpoint \(0\)
Explanation
\(f\) is continuous on \((0,1]\), but the interval is not closed; near \(0\) the values grow without bound and there is no maximum.
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
- ZW-EXTREME-VALUE-THEOREM-001 For which of these functions is an absolute maximum NOT guaranteed by the extreme value theorem?
- ZW-EXTREME-VALUE-THEOREM-002 Which condition of the extreme value theorem fails for f(x)=(1)/(x) on [-1,1]?
- ZW-EXTREME-VALUE-THEOREM-003 What does the extreme value theorem guarantee for a function continuous on [a,b]?
- ZW-EXTREME-VALUE-THEOREM-004 Which of these intervals is closed and bounded?
- ZW-EXTREME-VALUE-THEOREM-005 What does it mean for f to have an absolute maximum on [a,b]?
- ZW-EXTREME-VALUE-THEOREM-006 Which of these functions is NOT continuous on [0,1]?
- ZW-EXTREME-VALUE-THEOREM-007 Let f(x)=x^3-3x on [-2,2]. What is the absolute maximum value of f on that interval?
- ZW-EXTREME-VALUE-THEOREM-008 Which of these functions shows that the theorem fails when the function is not continuous?
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