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'Least of All' printed from https://nrich.maths.org/
Thank you Ruth from Manchester High School
for Girls for your solution to this problem.
Most people's first reaction to this
question is exactly the same as Ruth's:
"My initial conjecture was that the minimum value of $f(x)$ is when
$x=0$ for any value of $a$, because the function is even and
increases as $|x|$ increases."
Using calculus, we shall see that this is not so and that the
minimum value of $f(x)$ does depend on the value of $a$ .
The given expression is \begin{eqnarray} f(x) &=& (1 +
(a+x)^2)(1 + (a-x)^2)\\ &=&(1 + x^2 + a^2 +2ax)(1 + x^2
+a^2 -2ax)\\ &=& (1 + x^2 + a^2)^2 - 4a^2x^2 \\ &=&
x^4 + 2x^2(1 + a^2) + (1 +a^2)^2 - 4a^2x^2 \\ &=& x^4 +
2x^2(1 - a^2) + (1+a^2)^2. \end{eqnarray}
As this is a quartic in $x$ there will be one or three turning
points.
Differentiating $f$ to find the minima:
$$f'(x) = 4x^3 + 4x(1-a^2) = 4x(x^2 + (1 - a^2))$$
Case 1 : $(1 - a^2) <
0$.
The derivative $f'(x) = 0$ for $x = 0$ and $x = \pm \sqrt (a^2 -
1)$. The second derivative $f''(x) = 12x^2 + 4(1 - a^2) < 0$ at
$x = 0$ which gives a maximum value but $f''(x) = 12x^2 + 4(1 -
a^2) > 0$ for $x = \pm \sqrt (a^2 - 1)$ giving two minimum
points on the quartic where $x = \pm \sqrt (a^2 - 1)$. The minimum
value at each point is $f(x) = 4a^2$ where the position of X for
these minimum values is clearly dependent of $a$.
Case 2 : $(1 - a^2) >
0$.
The derivative $f'(x) = 0$ if and only if $x = 0$. The second
derivative
$f''(x) = 12x^2 + 4(1 - a^2) > 0$ so there is one minimum value
$f(x)= (1+a^2)^2$ where the position of X, at $x=0$, is independent
of $a$ agreeing with the conjecture.
Case 3 : $(1 - a^2) =
0$.
Note that where $a^2 = 1$ 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 2 but does not
allow the possibility of Case 1. Realising that the function we are
minimizing is a quartic we should have taken into account the
possibility of two minimum values.