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

Let \(f(x)=\lfloor x\rfloor\) (floor function) on \([0,2]\), with \(f(0)=0\) and \(f(2)=2\). It never equals \(\tfrac12\). Why does this not contradict the theorem?

Choose one option.

Show answer
  • C. \(f\) is not continuous on \([0,2]\): it jumps at \(x=1\) and \(x=2\)

Explanation

\(\tfrac12\in(0,2)\), so the value is between the endpoints; what fails is continuity, since \(\lfloor x\rfloor\) has jumps.

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Review the concept: Intermediate value theorem

If \(f\) is continuous on \([a,b]\) and \(N\) lies between \(f(a)\) and \(f(b)\), there is some \(c\in(a,b)\) with \(f(c)=N\).

\[f\ \text{continuous on }[a,b],\ f(a)

More practice problems: intermediate value theorem

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