([KRD10, Exercise 6.2.B]) If \(f:(a,b)\to\mathbb R\) is continuously differentiable and \(f'(x_0)\ne0\) for some \(x_0\in(a,b)\), then \(f\) is injective on some interval \((c,d)\) containing \(x_0\).
Solution
1. \(f'\) keeps the sign of \(f'(x_0)\) in a neighborhood.Proof: assume \(f'(x_0) > 0\) (the case \(< 0\) is symmetric).Since \(f'\) is continuous at \(x_0\), there is \(\delta > 0\) with \(f'(x) > f'(x_0)/2 > 0\) for all \(x \in (c, d) := (x_0 - \delta, x_0 + \delta) \subseteq (a,b)\).
2. \(f\) is strictly increasing on \((c,d)\).Proof: for \(c < u < v < d\), by MVT there is \(\xi \in (u,v)\) with \(f(v) - f(u) = f'(\xi)(v - u) > 0\) (as \(f'(\xi) > 0\) and \(v - u > 0\)). Hence \(f(u) < f(v)\).
3. \(f\) is injective on \((c,d)\).Proof: a strictly increasing function is injective: if \(f(u) = f(v)\) with \(u \ne v\), say \(u < v\), then \(f(u) < f(v)\), contradiction.
4. Q.E.D.