Exercise · ZW-INTERMEDIATE-VALUE-THEOREM-011 · ★☆☆ Basic

Let \(f(x)=x^3-x-1\) on \([1,2]\). Since \(f(1)=-1\) and \(f(2)=5\), what does the intermediate value theorem guarantee?

Choose one option.

Show answer
  • B. That there is \(c\in[1,2]\) with \(f(c)=0\)

Explanation

Zero lies between \(-1\) and \(5\), so some \(c\) attains it; the theorem does not say where it is or that \(f\) is monotone, and \(7>5\) is out of range.

Study Calculus in ZWMath →

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

← ZW-INTERMEDIATE-VALUE-THEOREM-010 ZW-INTERMEDIATE-VALUE-THEOREM-012 →