Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. (more details to come)

Question

anonymous · Answer

$$f(x)=(x-7)/(x+3) and g(x)=(-3x-7)/(x-1)$$

anonymous · Answer

@hero Any ideas? I'm not really sure how to go about this... unless you're supposed to turn it into a complex fraction?

hero · Answer

$$f(x) = \frac{x - 7}{x + 3}$$
$$g(x) = -\frac{3x+7}{x - 1}$$

hero · Answer

Notice that each fractional expression is improper. What you do is re-write each fraction as a mixed fraction.

hero · Answer

When you do this, the resulting expression will have only one x variable. This is will make it easier for you to do f(g(x)) and g(f(x)).

hero · Answer

To convert f(x) to a mixed fraction:

$f(x)=\dfrac{x - 7}{x + 3} $

        $= \dfrac{x + 3 - 10}{x + 3}$

        $=\dfrac{x + 3}{x + 3} - \dfrac{10}{x + 3}$

        $=1 - \dfrac{10}{x + 3}$

hero · Answer

Notice that f(x) is now expressed as a mixed fraction with only one $x$ variable as input

hero · Answer

Now to convert g(x) to a mixed fraction, we will do a similar process:

hero · Answer

$$g(x) = -\frac{3x+7}{x - 1}$$
       $= -\dfrac{3x - 3  + 10}{x - 1}$

       $=-\dfrac{3x - 3}{x - 1} - \dfrac{10}{x - 1}$

       $=-3 - \dfrac{10}{x - 1}$

hero · Answer

So now, 

$f(g(x)) = 1 - \dfrac{10}{-3-\frac{10}{x-1} + 3}$

            $= 1 - \dfrac{10}{-3 + 3-\dfrac{10}{x-1}}$

            $=1 - \dfrac{10}{-\dfrac{10}{x-1}}$

            $=1 + \dfrac{10}{\dfrac{10}{x-1}}$
         
            $=1 + 10 \times \dfrac{x - 1}{10}$
         
            $=1 + x - 1$

            $=1 - 1 + x$

            $=x$

hero · Answer

I hope you are able to understand how I got that.

hero · Answer

Do a similar process for g(f(x)) to arrive at x

anonymous · Answer

thank you!

hero · Answer

Did you understand any of that?

hero · Answer

If you have any questions, now is the time to ask. I won't be around much longer.

anonymous · Answer

Yeah, I ended up with the correct answer.

hero · Answer

Did you calculate g(f(x)) yet?

hero · Answer

@Precalcstuent2013, it is possible that you may be having trouble with this. I skipped some steps just for the sake of posting a solution in a timely manner. It is easy to confuse some of the steps. So if you are having trouble with it, please let me know so I can clarify things. There is no need to go on being frustrated about any of the steps if you are having trouble. The method I have shown you is not a popular one.