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        riddles >> putnam exam (pure math) >> Inverse
        
(Message started by: Lizzy on Mar 22^nd, 2006, 2:31am)

Title: Inverse
Post by Lizzy on Mar 22^nd, 2006, 2:31am

this seems waaay too ambigous :S

Determine conditions on the constants a,b,c and d so that the rational function

f(x) = (ax+b)/(cx+d) has an inverse. What is the domain of f?

Title: Re: Inverse
Post by Icarus on Mar 22^nd, 2006, 6:06pm

Not really. It is fairly straight-forward. Let y = (ax+b)/(cx+d), then solve for x in terms of y. For what values of a, b, c, and d is it possible to calculate the resulting function x = g(y) for at least some values of y?

As for the domain, it is just all the values of x for which the function f(x) has a calculable value. The answer to both is found by asking the question, what numbers can you NOT divide by?

Title: Re: Inverse
Post by Lizzy on Mar 23^rd, 2006, 7:15pm

hmm, ok i got

y(cx+d)/a = x

now the only constraint i can see is that a cannot equal zero.

For every other value of y, ther should be a value of x.

Does that mean the domain of f, is simply everything except a= 0??

or am i totally wrong?

Title: Re: Inverse
Post by towr on Mar 24^th, 2006, 12:19am

You lost a 'b' somewhere, and you still have an x on both sides.

y = (ax+b)/(cx+d)
y*(cx+d) = ax+b
cxy+dy = ax+b
cxy-ax = b-dy
x*(cy-a) = b-dy
x = (b-dy)/(cy-a)

Title: Re: Inverse
Post by Icarus on Mar 24^th, 2006, 3:43pm

In addition to towr's comments about the algebra,

on 03/23/06 at 19:15:38, Lizzy wrote:

Does that mean the domain of f, is simply everything except a= 0??

The Domain of f has nothing to do with your calculation of an inverse. f(x) is a function of x, not of a (a, b, c, d are considered to be fixed values, while x varies). The domain of f is the set of all values of x (not a!) for which you can calculate a value for f(x) (not its inverse!).

I notice that I made a typo in my previous post. I originally had "f(y)", not f(x). I apologise if this confused you.

Perhaps a more concrete example would help to clarify the general formula. Set a=2, b=-1, c=1, d=2. So we are considering the function

f(x) = (2x-1)/(x+2).

What values can x take on, and still allow you to calculate a value for f(x)?

To figure it out, ask the opposite: for what values of x can you NOT calculate a value for f(x)? The only thing that can cause trouble is the division, since the denominator cannot be zero. Therefore we cannot have x + 2 = 0, or x = -2.

Hence the domain of this example is the set of all real numbers other than -2: { x | x != -2}.

Now, you do it for arbitrary a, b, c, d.

--------------------------------------------------------------

For the inverse problem, on the other hand, you need to figure out which values of a, b, c, d, prevent the calculation that towr has demonstrated from working. In particular, what values of a, b, c, d prevent (b-dy)/(cy-a) from existing for every possible value of y?

Title: Re: Inverse
Post by Lizzy on Mar 25^th, 2006, 4:41am

thanks for ur help! that made alot more sense that time.

Worked it out i think:P