As X moves along the line segment from P to Q, the distance XP
increases and the distance XQ decreases symmetrically. It seems possible
therefore that the minimum value will occur at the midpoint of XP and
that this will be independent of the length of the line segment, that is
independent of a. We shall see that this is not so. The given expression is
f(x)
=
(1 + (a+x)2)(1 + (a-x)2)
=
(1 + x2 + a2 +2ax)(1 + x2 +a2 -2ax)
=
(1 + x2 + a2)2 - 4a2x2
=
x4 + 2x2(1 + a2) + a4 - 4a2x2
=
x4 + 2x2(1 - a2) + (1+a2)2
Note that this is a quartic in x with the coefficient of x4 positive
and so we should have expected one or three turning points.
Differentiating f to find the minima:
f¢(x) = 4x3 + 4x(1-a2) = 4x(x2 + (1 - a2))
Case 1: (1 - a2) > 0
The derivative f¢(x) = 0 if and only if x = 0 and
the second derivative f¢¢(x) = 12x2 + 4(1 - a2) > 0.
So there is one minimum value f(x) = (1+a2)2 where the position of X, at
x=0, is independent of a. Case 2: (1 - a2) < 0
The derivative f¢(x) = 0 for x = 0 and x = ±Ö(a2 - 1).
The second derivative f¢¢(x) = 12x2 + 4(1 - a2) < 0 at x = 0 which
now gives a maximum value but f¢¢(x) = 12x2 + 4(1 - a2) > 0 for
x = ±Ö(a2 - 1) giving two minimum points on the quartic where
x = ±Ö(a2 - 1). The minimum value at each point is
f(x) = 4a2 where the position of X for these minimum values is
clearly dependent of a.
Case 3: (1 - a2) = 0
Note that where a2 = 1 there is a single minimum f(x) = 4 at
x=0 giving continuity between Case 1 and Case 2.
The most likely first conjecture agrees with Case 1 but does not allow the
possibility of Case 2. Realising that the function we are minimizing is a
quartic we should have taken into account the possibility of two minimum
values.