Question

Prove that g(x)=2x+3/x-1 is the inverse of f(x)=x+3/x-2. I found that it is the inverse just having trouble proving it. I already plugged the function in as x but am having trouble.

Answer 1

If you're going to write the inverse of a function, you should express the second function as g(x)

Answer 2

let's let the inverse function be g(x)

Answer 3

Rewrite g(x) as \[g(x)=1 + \frac{5}{x - 2}\]

Answer 4

You need to plug the second function as x in the first function, then simplify it;
after that, plug the first fuction as x in the second function, then simplify it
if both answer are x then you can say that f(x) = g(x) = x therefore they are inverse

Answer 5

Then find g(f(x))

Answer 6

as @ktrinh2296 pointed out, if g(x) is truly the inverse of f(x), then f(g(x)) should equal x.

Answer 7

\[\frac{x + 3}{x - 2} = 1 + \frac{5}{x - 2}\]

Answer 8

I meant to let that second function equal g(x)

Answer 9

If we do that then to find g(f(x)), we do this:

\[g(f(x)) = 1 + \frac{5}{\frac{2x + 3}{x - 1} -2}\]

Answer 10

how did you get 1+(5/x-2)

Answer 11

First, before I show you that, I will show you that g(f(x)) = x

Answer 12

ok

Answer 13

\[g(f(x)) = 1 + 5 \div \left(\frac{2x + 3}{x - 1} - 2\right)\]
\[ = 1 + 5 \div \left(\frac{2x + 3 - 2(x - 1)}{x -1}\right)\]
\[ = 1 + 5 \div \left(\frac{2x + 3 -2x + 2}{x -1}\right)\]
\[ = 1 + 5 \div \left(\frac{2x -2x+ 3 + 2}{x -1}\right)\]
\[ = 1 + 5 \div \left(\frac{5}{x -1}\right)\]
\[ = 1 + 5 \times \left(\frac{x-1}{5}\right)\]
\[ = 1 + \left(\frac{5(x-1)}{5}\right)\]
\[ = 1 + \left(\frac{\cancel{5}(x-1)}{\cancel{5}}\right)\]
\[ = 1 + (x - 1)\]
\[ = 1 + x - 1\]
\[ = x + 1 - 1\]
\[ = x\]

Answer 14

oh ok

Answer 15

Now, as promised:

\(\large\frac{x + 3}{x - 2} \)
\[\\= \frac{x - 2 + 5}{x - 2}
\\= \frac{x - 2}{x - 2} + \frac{5}{x - 2}\]
\[= 1 + \frac{5}{x-2}\]

Answer 16

ohh ok thank you a bunch. I gave you a metal. If I could have I would have given you 3. :)

Answer 17

I understand it perfect

Answer 18

You're welcome

Answer 19

Rewriting g(f(x)) as a whole number plus a fraction makes life much easier when trying to find the composition of functions.

Answer 20

Yes, that is for sure. That is where I made a mistake.

Answer 21

It wasn't a mistake, you probably just never knew about it. But for reference you should know that something like

\[\frac{x + 3}{x -2}\] is actually an improper fraction so it can be re-written as a mixed number.

Answer 22

If this is homework you have to turn in tomorrow, you'll probably have the best solution in the class.

Answer 23

But for reference, we let the first function be f(x) and the second function be g(x). As it is, now, you have it backwards.

Answer 24

Just started Calculus. So that should help me. Thanks a lot again.

Answer 25

It will definitely help you in case you have to integrate a fraction in such form.

Answer 26

Ok again it helped a lot.