> [!theorem] > > Let $f: [a, b] \to \real$ be a [[Derivative|differentiable]] [[Function|function]]. If $f^\prime(a) < \alpha < f^\prime(b)$, then $\exists c \in (a, b)$ where the [[Derivative|derivative]] $f^\prime(c) = \alpha$. > > *Proof*. Consider the function $g(x) = f(x) - \alpha x$, then $g$ is differentiable on $[a, b]$ with $g^\prime = f^\prime - \alpha$. which makes > $ > f^\prime(a) < \alpha < f^\prime(b) \Leftrightarrow g^\prime(a) < 0 < g^\prime(b) > $ > and $g^\prime(c) = 0$ the new target. First, since $g^\prime(a) < 0$, > $ > \exists \delta > 0: |x - a| < \delta \Rightarrow \frac{g(x) - g(a)}{x - a} < 0 > $ > and therefore $\exists x \in (a, b): g(a) > g(x)$. Similarly, $\exists y \in (a, b)$ with $g(y) < g(b)$. Recursively construct an increasing $x$ sequence and a decreasing $y$ sequence, which both converge ([[Monotone Convergence Theorem for Sequences|monotone convergence theorem]]) to somewhere in the pre-image of the minimum, at which point $g^\prime$ would be $0$ ([[Interior Extremum Theorem]]).