function for the
parabola y = f(x) = -x2 + 10x +3, you
know that it has a maximum value which can be found by either
completing the square or, in calculus, taking the derivative.
If we assume that the extreme value of a parabola lies on the axis, we can
use the following method of determining the axis of symmetry to find the maximum value.
Write the function as y = x(10 - x) + 3. 10 - x is a
self-invere function. Because of this,
f(10 - x) = (10 - x) x
+ 3 = f(x). For all x, f(10 - x) = f(x)
For any value x, we can find the other value, 10
- x, that has the same value for f. Another way of saying this is that for every point
(x,y) on the parabola there is a point (10 - x, y). The midpoint of (x,y) and (10 - x,y) is
( (x+(10-x))/2, y) = (5,y). The line x=5 is a line of ymmetry for the parabola.
The maximum value is for x = 5. f(5) = 28, so the maximum occurs at (5, 28).
The method also works if the coefficient of the x2 is 1 instead of -1.
For example, consider f(x) = x2
+ 10x + 3. We know the parabola in this case has a minimum value. f(x)
can be rewritten as f(x) = -x(-10 - x) +3. f(-10 - x) = f(x). The axis
of the parabola is the line x = -5.
Now look at the equation y = x2/(x - 1).
There is a discontinuity at x = 1, so the function divides
into two pieces for x < 1 and x > 1. Let's look at the
portion for x > 1. It should be clear that for x near 1 and for
large values of x, the function goes to infinity, so we can suppose
that it takes on a minimum value in between.
I came up with this function by taking the self-inverse
function x/(x - 1) above and multiplying it by x. That is,
f(x) = x(x/(x-1)). We can use this construction as in the case of the
parabola to show that f(x/(x - 1)) = (x/(x - 1))x =
f(x). If we assume that the function for x < 1 takes a
minimum for only one value of x, then arguing as in the case for the
parabola, at the minimum we must have x = x/(x - 1).
Muliplying both sides by x - 1 we get x2 - x = x
, x2 - 2x = 0, x(x - 2) = 0. For x
> 1, the only solution is x = 2, for which f(x) = 4.
Looking at the graph of the function it can be seen that we
were correct in assuming that the function takes on a minimum at only
one place and that the minimum does indeed occur for x = 2. The other
solution for x(x - 2) =0, x = 0, is a maximum for the portion of the
graph where x < 1.