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.
can generalize what we did with the parabola. Suppose we have f(x) =
arctan(x) + arctan(4 - x). Find the line of symmetry. This
may look scary, but it is easy to solve doing what we did for the
paraboloa. We have f(4 -x) = arctan(4-x) + arctan(4 - (4-x)) =
arctam(4-x) + arctan(x) = f(x). f(4-x) = f(x). Just as we
did above, we can say that there is an axis of symmetry for the line x
= 2. For any function g(x), f(x) = g(x) + g(a -x) has a line of
symmetry at x=a/2. This will hold for f(x)=g(x)g(a-x) or any
other function that can be written in terms of g(x) and g(a-x) which is
symmetric in g(x) and g(a-x).
Suppose that we had a
function f(x) = g(a-x) + g(b+x). What can we do with this?
Let's look at f(x-b). This moves the curve to the right by
b and will get rid of the b value. We have f(x-b) = g(a-(x-b)) +
g(b+(x-b)) = g((a+b)-x) + g(x). This is in the form we want.
There willl be an axis of symmetry for f(x-b) at x = (a+b)/2. To
get to our original function, we move to the left by b, which will move
the axis of symmetry to the left as well. The axis of symmetry
for f(x) is (a+b)/2 - b = (a-b)/2. Based on what we did before,
going in the reverse direction suggests that f(a-b - x) = f(x). I
leave it as an exercise to show from the definition of f(x), this is
indeed the case.. If I had been more astute, I would have noticed it
right away, without having to first look at f(x-b).
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.