# Re: pre calculus inverse functions

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Posted by T.Gracken on September 03, 2002 at 16:29:58:

In Reply to: pre calculus inverse functions posted by Teena on September 03, 2002 at 13:55:20:

: f(x) = 4x/x-2

: how do i solve for x ?

If f(x) = 4x/x - 2, then f(x) = 4 - 2 = 2.

this means Mr.S. is correct and the function cannot be solved in terms of x.

But...

if you mean f(x) = (4x)/(x-2), then

1. multiply both sides by (x-2) to get

f(x)*(x-2) = 4x

2. distribute:

x*f(x) - 2*f(x) = 4x

3. isolate terms in x

x*f(x) - 4x = 2*f(x)

4. factor x:

x*[f(x) - 4] = 2*f(x)

5. divide both sides by coefficient of x:

x = [2*f(x)] / [f(x) - 4]

You now have a function in two variables (x and f(x)) solved for the domain variable (a.k.a. abscissa), where they are generally solved for the range variable (a.k.a. ordinate).

I would suggest (for easier reading) when you want to solve a function (in x) for x, rewrite the ordinate [that is, the f(x)'s] as y's.

note: in this case, x can not equal 2. in the first case, x can not equal 0. (...why???)

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