# Inverse and composition of functions

1. Nov 10, 2015

### terryds

1. The problem statement, all variables and given/known data

If $f(2x-1)= 6x + 15$ and $g(3x+1)=\frac{2x-1}{3x-5}$, then what is $f^{-1}\circ g^{-1}(3)$ ?

a) -2
b) -3
c) -4
d) -5
e) -6

3. The attempt at a solution

I think the f inverse and g inverse is
$f^{-1}(6x+15)= 2x-1$
$g^{-1}(\frac{2x-1}{3x-5})=3x+1$

and,$f^{-1}\circ g^{-1}(3)=f^{-1}(g^{-1}(3))$

Then,

I equate $\frac{2x-1}{3x-5}=3$
So, I get x = 2

So, $g^{-1}(3)=3(2)+1=7$

Then, I equate $6x+15=7$
And, I get x = 8/6 or 4/3

So, $f^{-1}(7)=2(\frac{4}{3})-1=\frac{5}{3}$ which does not appear in the options..

2. Nov 10, 2015

### PWiz

Let's deal with $g$ first. Let the input ($3x+1$) be equal to $u$. Then $x=\frac{u-1}{3}$. So you have g(u) = something, and you have to find that something by expressing the output in terms of $u$. Then try inverting the function.

3. Nov 10, 2015

### terryds

I get $g(u)=\frac{2u-5}{3u-18}$
Then $g^{-1}(x)=\frac{18x-5}{3x-2}$
So, $g^{-1}(3)=7$

Then it'll be $f^{-1}(7)=\frac{5}{3}$
The answer is -5 according to the book.. But, it doesn't give me the explanation..

4. Nov 10, 2015

### Samy_A

That should be $x=-4/3$. Still doesn't give the answer from the book though.

5. Nov 10, 2015

### terryds

Yup, perhaps the question is wrong

6. Nov 10, 2015

### Samy_A

Or I'm missing something. I redid your calculation, and also did it like @PWiz suggested (for the two functions), but with both methods I get $-11/3$ as result.

7. Nov 10, 2015

### PWiz

I just did the calculation myself. I'm getting -11/3 as the answer. I'm guessing something is wrong with the answer given.
EDIT: I see that @Samy_A has gotten the same result. That pretty much confirms that something is wrong with the question, since we both can't get the same wrong answer.