Exercise 1274Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

Rolle and the sign of $f'$

#Rolle#sign change#derivative

Statement

Let $f$ be continuous on $[a,b]$ , differentiable on $]a,b[$ , and $f(a)=f(b)=0$ with $f\not\equiv0$ . Show that there exists $c\in]a,b[$ with $f'(c)\neq0$ .

Hint

If $f'(x)=0$ everywhere, then $f$ is constant (consequence of MVT); but $f\not\equiv0$ .

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Solution

Contradiction. If $f'(x)=0$ for all $x$ , then by the MVT $f$ is constant on $[a,b]$ , contradicting $f\not\equiv0$ .

Conclusion. There exists $c\in]a,b[$ with $f'(c)\neq0$ .

Rolle and the sign of $f'$

Statement

Solution

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