Suppose that \(f,g:\mathbb{R}\to\mathbb{R}\) are differentiable, that \(f(x)\leq g(x)\) for all \(x\in\mathbb{R}\), and that \(f(x_0)=g(x_0)\) for some \(x_0\). Prove that \(f'(x_0)=g'(x_0)\).
Solution
1. Reduce to a non-negative function with a minimum at \(x_0\).
set \(h = g - f\); then \(h \ge 0\), \(h(x_0) = 0\), and \(h\) is differentiable with \(h' = g' - f'\).
2. Right-hand difference quotients of \(h\) at \(x_0\) are non-negative.
for \(x > x_0\), \(h(x) \ge 0 = h(x_0)\), so \((h(x) - h(x_0))/(x - x_0) \ge 0\); hence the right derivative \(h'_+(x_0) = \lim_{x \to x_0^+}\frac{h(x)-h(x_0)}{x-x_0} \ge 0\).
3. Left-hand difference quotients are non-positive.
for \(x < x_0\), \(x - x_0 < 0\) and \(h(x) - h(x_0) \ge 0\), so \((h(x)-h(x_0))/(x-x_0) \le 0\); hence \(h'_-(x_0) \le 0\).
4. \(h'(x_0) = 0\), hence \(f'(x_0) = g'(x_0)\).
\(h\) is differentiable at \(x_0\), so \(h'(x_0) = h'_+(x_0) = h'_-(x_0)\); by 2 and 3 this common value satisfies \(0 \le h'(x_0) \le 0\), so \(h'(x_0) = 0\).
5. Q.E.D.