Question 7.T.12: (Darboux) Let the function f : [a, b] → R be differentiable....
(Darboux)
Let the function f : [a, b] → \mathbb{R} be differentiable. If λ is a number between f^{′}(a) and f^{′}(b), then there is a point c ∈ (a, b) such that f^{′}(c) = λ.
By using −f instead of f if necessary, we may assume that
f^{′}(a) < λ < f^{′}(b). (7.9)
The function defined by
g(x) = f(x) − λx for all x ∈ [a, b]
is differentiable on [a, b], and
g^{′}(x) = f^{′}(x) − λ for all x ∈ [a, b].
If there is no c ∈ (a, b) such that f^{′}(c) = λ, then, for every x ∈ (a, b), g^{′}(x) ≠ 0. By Theorem 7.8, this means that g is injective on [a, b]. Since g is also continuous, it is strictly monotonic (by Theorem 6.7). If g is strictly increasing, then
\frac{g(x) − g(a)}{x − a} > 0 for all x ∈ (a, b),
hence
g^{′}(a) = \underset{x→a}{\lim } \frac{g(x) − g(a)}{x − a} ≥ 0.
This contradicts the assumption (7.9) that g^{′}(a) = f^{′}(a) − λ < 0. If g, on the other hand, is strictly decreasing, then
\frac{g(x) − g(a)}{x − a} < 0 for all x ∈ (a, b),
and we obtain g^{′}(b) ≤ 0 in the limit as x → b, thereby contradicting the assumption that g^{′}(b) = f^{′}(b) − λ > 0. Thus there is a c ∈ (a, b) where g^{′}(c) = f^{′}(c) − λ = 0.