Problem 11892
(American Mathematical Monthly, Vol.123, February 2016) Proposed by F. Perdomo and A. Plaza (Spain).
Letf be a real-valued continuously differentiable function on [a, b] with positive derivative on (a, b).
Prove that, for all pairs(x1, x2) with a ≤ x1< x2≤b and f (x1)f (x2) > 0, there exists t ∈ (x1, x2) such that
x1f (x2) − x2f (x1)
f (x2) − f (x1) = t − f (t) f′(t).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
The function f is increasing in [a, b], and f (x1)f (x2) > 0 implies that f (x) 6= 0 in [x1, x2] (otherwise f (x1) ≤ 0 ≤ f (x2)). Let F (x) := −x/f (x) and G(x) := −1/f (x), then F and G are both continuous in [x1, x2], and differentiable in (x1, x2). Moreover G′(x) = f′(x)/f2(x) 6= 0 in (x1, x2). Then, by the Cauchy’s Mean Value Theorem there exists some t ∈ (x1, x2), such that
x1f (x2) − x2f (x1)
f (x2) − f (x1) = −f(xx2
2)+f(xx1
1)
−f(x1
2)+f(x1
1)
=F (x2) − F (x1)
G(x2) − G(x1) =F′(t) G′(t) =
−f (t)+tf′(t) f2(t) f′(t) f2(t)
= t − f (t) f′(t).
