Answer to Question #230895 in Calculus for Samuel

Question

Let f be differentiable on R. Suppose f'(x) not equal 0 for every x. Prove that f has at most one real root

Answer 1

Suppose to the contrary that $f$ has at least two real roots at $x=a$ and $x=b,$ where $a<b.$

Then, $f$  is by hypothesis continuous on $[a, b]$ and differentiable on $(a, b)$ such that $f(a)=0=f(b).$

By the Rolle's Theorem there is a number $c\in(a, b)$ such that $f'(c)=0.$

This contradicts the fact that $f'(x)\not=0$ on $\R.$

Therefore, we conclude that $f$ has at most one real root.