(June 1999 #10) Show that if \(f\) is differentiable on \((a,b)\) with \(f'(x)=0\) on \((a,b)\), then \(f\) is constant on \((a,b)\).
Solution
1. Fix \(x < y\) in \((a,b)\) and apply the mean value theorem.Proof: \(f\) is differentiable on \((a,b)\), hence continuous on \([x,y]\) and differentiable on \((x,y)\).By MVT there is \(\xi \in (x,y)\) with \[f(y) - f(x) = f'(\xi)(y - x) = 0\cdot(y-x) = 0,\] since \(f' \equiv 0\) on \((a,b)\).
2. Conclude.Proof: \(f(x) = f(y)\) for all \(a < x < y < b\); hence \(f\) is constant on \((a,b)\).
3. Q.E.D.