514 Real Analysis
namely,x =^23 andx =−^12. We computeg(− 1 ) =4,g(−^12 )=13,g(^23 ) = 278 ,
g( 1 )=4. The largest of them is 13, which is therefore the answer to the problem. It is
attained whenz=−^12 ±
√ 3
2 i.
(8th W.L. Putnam Mathematical Competition, 1947)408.After we bring the function into the form
f(x)=(
x− 1 +x^1) 3
x^3 − 1 +x^13,
the substitutionx+x^1 =sbecomes natural. We are to find the minimum of the function
h(s)=
(s− 1 )^3
s^3 − 3 s− 1= 1 +
− 3 s^2 + 6 s
s^3 − 3 s− 1over the domain(−∞,− 2 ]∪[ 2 ,∞). Setting the first derivative equal to zero yields the
equation
3 (s− 1 )(s^3 − 3 s^2 + 2 )= 0.The roots ares=1 (double root) ands= 1 ±
√
- Of these, onlys= 1 +
√
3 lies in the
domain of the function.
We compute
lim
x→±∞
h(s)= 1 ,h( 2 )= 1 ,h(− 2 )= 9 ,h( 1 +√
3 )=
√
3
2 +
√
3
.
Of these the last is the least. Hence the minimum offis
√
3 /( 2 +
√
3 ), which is attained
whenx+^1 x= 1 +
√
3, that is, whenx=( 1 +√
3 ±^4
√
12 )/2.
(Mathematical Reflections, proposed by T. Andreescu)409.Letf(x)=sin(sin(sin(sin(sin(x))))).The first solution isx=0. We have
f′( 0 )=cos 0 cos(sin 0)cos(sin(sin 0))cos(sin(sin(sin 0)))cos(sin(sin(sin(sin 0))))= 1 >1
3
.
Therefore,f(x) >x 3 in some neighborhood of 0.On the other hand,f(x) <1, whereas
x
3 is not bounded asx→∞. Therefore,f(x^0 )=
x 0
3 for somex^0 >^0 .Becausefis odd,
−x 0 is also a solution. The second derivative offis
−cos(sinx)cos(sin(sinx))cos(sin(sin(sinx)))cos(sin(sin(sin(sinx))))sinx
−cos^2 xcos(sin(sinx))cos(sin(sin(sinx)))cos(sin(sin(sin(sinx))))sin(sinx)