Exercise 616Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

A function with zero derivative is constant

#consequence#constant function#zero derivative

Statement

Show that if $f$ is continuous on $[a,b]$ , differentiable on $]a,b[$ , and $f'(x)=0$ for all $x\in]a,b[$ , then $f$ is constant on $[a,b]$ .

Hint

Apply the mean value theorem to any pair $x<y$ .

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MVT. For any $x<y$ in $[a,b]$ , there is $c\in]x,y[$ with $f(y)-f(x)=f'(c)(y-x)=0$ , so $f(x)=f(y)$ .

Conclusion. $f$ is constant.