consider the following function

$\displaystyle f(x)=\frac{x}{x+5}$

find the inverse function. express it as a function of x.

The answer is $\displaystyle f^{-1}(x)=\frac{5x}{1-x}$

but what is the process to getting that answer???

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- Oct 16th 2011, 07:21 PMdelgeezeehelp finding the inverse
consider the following function

$\displaystyle f(x)=\frac{x}{x+5}$

find the inverse function. express it as a function of x.

The answer is $\displaystyle f^{-1}(x)=\frac{5x}{1-x}$

but what is the process to getting that answer??? - Oct 16th 2011, 07:38 PMSammySRe: help finding the inverse
There are several ways to do this.

I think the one common in College Algebra class is:

For y = f(x), you have $\displaystyle y=\frac{5x}{1-x}$

To find $\displaystyle y = f^{-1}(x)$ interchange all x's & y's, i.e., wherever there was an x place a y & wherever a y (usually just the one) place a x. Then solve for y. What you then have for y is the same as what is meant by $\displaystyle f^{-1}(x)$. In your case solve $\displaystyle x=\frac{5y}{1-y}$ for y and that's the same as the inverse function. - Oct 16th 2011, 07:39 PMTKHunnyRe: help finding the inverse
In the casual language, it's "Swap and Solve"!

If y = x/(x+5), the inverse, if it exists, should be x = y/(y+5), but no one likes x interms of y, so re-solve for y.

xy + 5x = y

5x = y - xy = y(1-x)

y = 5x/(1-x)

It is maybe a little more obvious without the function notation.

Note: This is a VERY beginning concept in teh area of the function inverse. One cannot always solve after swapping. There are reflections to think about and Domain and Range issues. "Swap and Solve" is just the VERY beginning.