Exercise · ZW-INTERMEDIATE-VALUE-THEOREM-020 · ★★★ Advanced

How many real roots does \(f(x)=x^3-3x\) have on \([-2,2]\), where \(f(-2)=-2\) and \(f(2)=2\)?

Choose one option.

Show answer
  • A. \(3\)

Explanation

\(x^3-3x=x(x^2-3)=0\) gives \(x=0,\pm\sqrt3\), and all three lie in \([-2,2]\). The theorem only ensures at least one.

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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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