SOLUTION: Given the rational function f(x) = (x + 2)/(x - 2), find the inverse. Let me see. Let f(x) = y y = (x + 2)/(x - 2) x = (y + 2)/(y - 2) Stuck here....

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Question 1208167: Given the rational function f(x) = (x + 2)/(x - 2), find the inverse.

Let me see.

Let f(x) = y

y = (x + 2)/(x - 2)

x = (y + 2)/(y - 2)

Stuck here....
Found 5 solutions by josgarithmetic, math_tutor2020, greenestamps, MathTherapy, mccravyedwin:
Answer by josgarithmetic(39625)   (Show Source): You can put this solution on YOUR website!
To be brief or crude, switch x and y and that could be the inverse. Solve that for y.


Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

f(x) = (x+2)/(x-2)
y = (x+2)/(x-2)
x = (y+2)/(y-2) .... swap x and y; solve for y
x(y-2) = y+2
xy-2x = y+2
xy-y = 2+2x
y(x-1) = 2x+2
y = (2x+2)/(x-1)
g(x) = (2x+2)/(x-1)

g(x) is the inverse of f(x), and vice versa.
To verify, you need to show that,
f( g(x) ) = x and g( f(x) ) = x
for all x in the domain. I'll leave this verification for the student to do.
Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


The difficulty in finding the inverse in this example is because the variable x appears twice in the given function. For many complicated functions, that makes it impossible to find an inverse function.

For this function, which is a rational function with linear polynomials in the numerator and denominator, tutor @math_tutor_2020 shows you the algebra required to find the inverse function.

There are many types of problems in which it can be useful, given a rational function like this, to change the form of the function by "performing the division". For your example,



In the final form there, the variable x appears only once; you could find the inverse by the usual process of switching x and y and solving for the new y.

Unfortunately, you still end up having to do awkward algebra to find the inverse.

However, given a function in which the variable occurs only once, there is a very different way for finding the inverse function -- based on the concept that an inverse function "un-does" what the function does; stated differently, the inverse "gets you back where you started". For your example, finding the inverse is much easier by this method than by using the standard method, once we have rewritten the given function in a way that has the variable occurring only once.

To get you back where you started, the inverse function, compared to the given function, needs to perform the opposite operations in the opposite order.

So look at the function in its new form, , and determine the sequence of operations it performs on the input variable:
(1) subtract 2 (gives )
(2) take the reciprocal (gives )
(3) multiply by 4 (gives )
(4) add 1 (gives )

The inverse function must perform the opposite operations in the opposite order:

(1) subtract 1 (gives )
(2) divide by 4 (gives )
(3) take the reciprocal (gives )
(4) add 2 (gives )

So the inverse of the given function is

The inverse shown by the other tutors is

The two inverses are equivalent -- just in different forms:
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
Given the rational function f(x) = (x + 2)/(x - 2), find the inverse. 

Let me see.
Let f(x) = y
y = (x + 2)/(x - 2)
x = (y + 2)/(y - 2)
Stuck here....

You GOT/HAVE it!
        x =  <====== WHERE YOU ARE!
 x(y - 2) = y + 2 ------- Cross-multiplying 
  xy - 2x = y + 2
   xy - y = 2 + 2x
 y(x - 1) = 2 + 2x
        
 --- Replacing y with f- 1(x)
Answer by mccravyedwin(408)   (Show Source): You can put this solution on YOUR website!
Here's an easy way to find the inverse when both numerator and denominator are
linear:



The inverse of a rational function with a linear numerator and denominator is also a rational 
function with a linear numerator and denominator.

We know that f(x) has a vertical asymptote at x = 2 and a horizontal asymptote at y = 1,

So we know that f-1(x) has a vertical asymptote at x=1 and a horizontal asymptote at y = 2,
so we know the denominator could be , and if it were, the numerator would have to be
2x + something, so we have:

 

Now since f(x) goes through (0,-1), f-1 goes through (-1,0), so



and A has to be 2 

So .

You could just about do that in your head.

Edwin
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