For example, if f(x) = x for all x and g(x) = x for all x, then the domain of f/g is all x but x = 0 since g(0) = 0. Thus (f/g)(x) = 1 if x is non-zero and (f/g) is not defined at zero.

As another example if f(x) = {sqrt}(x) and g(x) = {sqrt}(x) then the domain of f and the domain of g are both the non-negative real numbers, so the domain of fg cannot contain any negative numbers. Thus (fg)(x) = {sqrt}(x){sqrt}(x) = {sqrt}(x{^2}) = |x| if x > 0 or x = 0, and (fg)(x) is not defined for negative numbers.

In Probem 6, we have (f/g)(x) = (x-1/x-2)/(x+1/x+2). The domain of f is all x other than 2 and the domain of g is all x other than -2. There is one value of x, namely -1, where g(x) = 0 and this also must be excluded from the domain of the quotient (f/g). Thus the domain of f/g is all numbers except for -1, 2, and -2, even though when we simplify algebraically we get (f/g)(x) = ((x-1)(x+2))/((x-2)(x+1)) which appeard to be all right at x = -2.

There are many ways to identify which functions belong to which
graphs in the list. Thus we can identify the graph of x/(x{^2}
- 9) by looking at the y-intercept and noting which graphs pass
through the point (0,0). We can also look for the graphs which
are not defined over the points x = 3 and x = -3. Either answer
is good, and there can be other answers as well.

Note
however that not all answers work. If the equation is cubic,
then the graph does not have to have three real roots. It might
have one, or three, or even two in case the graph has a local
maximum or minimum with a 0 y-coordinate. Compare the graphs of
x{^3} - 3x, x{^3} - 3x + 3 and x{^3} - 3x + 2. A similar
comment holds about a fifth degree equation. If there are more
than three roots, the equation has to have degree greater than
three and if there are five roots, the equation has to have
degree at least five. But not all fifth degree equations have
five roots, for example x{^5} - 1 which has x = 1 as its only
root.

More later.

Thomas Banchoff Last modified: Thu Sep 19 19:47:49 EDT 2002