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Hi!! I would really appreciate it if you could help me with Inverse Functions. I am supposed to prove that f and g are inverse functions but I seem to be missing a step along the way. I am also supposed to create a table which I have no idea how to do. Could you please help me solve a problem along with an explanation as well?... Here's one I need help on...

f(x)= -7/2x-3      g(x)= -2x+6/7

I would really appreciate your help.  

Answer
Questioner:   Katie
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  Inverse Functions
Question:  Hi!! I would really appreciate it if you could help me with Inverse Functions. I am supposed to prove that f and g are inverse functions but I seem to be missing a step along the way. I am also supposed to create a table which I have no idea how to do. Could you please help me solve a problem along with an explanation as well?... Here's one I need help on...

f(x)= -7/2x-3      g(x)= -2x+6/7

I would really appreciate your help.          
.......................................
1. There is no such thing as 'an inverse function' all by itself.

2. Two functions  f(x) and g(x) are 'inverses of each other' if both:

f( g(x)) = x

and

g( f(x)) = x

To find the inverse of some function, such as:

f(x) = 3x + 2    << I am not doing yours because I am not sure what you meant -- you did not parenthesize carefully.

Step 1:  WRITE  y = ....

y = 3x + 2

Step 2:  SWITCH x and y:

x = 3y + 2

Step 3: Do the ALGEBRA to solve for y:
     x - 2
y =  -------
       3

Step 4: MAKE UP a name for the function:

       x - 2
g(x) = -------
         3

Step 5:  CHECK:

f(g(x)) = 3 g(x) + 2
             x - 2
f(g(x)) = 3 (--------) + 2
               3
f(g(x)) = x - 2 + 2

f(g(x)) = x   << good.

Also:
         f(x) - 2
g(f(x)) = ----------
             3

         3x + 2 - 2
g(f(x)) = ----------
             3

          3x
g(f(x)) = -----
           3

g(f(x)) = x  << also good.

That's it. For that table stuff, you pick x's and compute y = f(x) for each, then you turn it around for the inverse.  

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