DANIEL ULLMAN, MONTHLY Problems
Department of Mathematics
The George Washington University
2201 G Street, NW, Room 428A
Washington, DC 20052
10739. Proposed by Occar Ciaurri, Logroño, Spain. Suppose that ƒ: [0, 1]   has a continuous
second derivative with ƒ''(x) > 0 on (0, 1), and suppose that ƒ(0) = 0. Choose
ƒ(b)
a  (0, 1) such that ƒ'(a) < ƒ(1). Show that there is a unique b  (a, 1) such that ƒ'(a) = b .
Proof.
1.)
If ƒ''(x) is continuous on (0, 1), then it is also integrable on (0, 1). Since ƒ''(x) > 0 on (0, 1)
x2
''
'
'
if we take x1 < x2  (0, 1) then 
x ƒ (t) dt = ƒ (x2) – ƒ (x1) > 0 
1

ƒ'(x2) > ƒ'(x1) , ie. that ƒ'(x) is strictly increasing on (0, 1).
2.)
Take b1  (0, a); by the Mean Value Theorem (henceforth MVT)  c  (0, b1 ) such that
ƒ(b1) – ƒ(0)
ƒ(b1)
'
'
=
b1 = ƒ (c) . Since c < b1 < a and ƒ (x) is strictly increasing on
b1 – 0
(0, 1), ƒ'(c) =
3.)
4.)
ƒ(b1)
ƒ(1) ƒ(x)
'(a) < ƒ(1) =
<
ƒ
b1
1 . x is continuous on (0, 1) so it has the
ƒ(b)
Intermediate Value Property   b  (b1 , 1) such that b = ƒ'(a) .
ƒ(b)
Suppose b ≤ a ; by the MVT  c1  (0, b) such that b = ƒ'(c1) which contradicts
that ƒ'(x) is strictly increasing on (0, 1). So b  (a, 1), as desired.
ƒ(b1)
ƒ(b2)
ƒ(b1)
ƒ(b2)
Suppose  b1 < b2  (a, 1) such that b
= b
= ƒ'(a) . b
= b

1
2
1
2
ƒ(b1) – ƒ(b2)
ƒ(bi)
'
=
bi = ƒ (a) , for i = 1, 2. By the MVT,  c2  (b1 , b2 ) such that
b1 – b2
ƒ'(c2) =
ƒ(b1) – ƒ(b2)
= ƒ'(a) , again contradicting that ƒ'(x) is strictly increasing on
b1 – b2
(0, 1). Consequently, b is unique, as desired.
Allen J. Mauney
1215 Kaiser Road NW
QED
Olympia, WA 98502