# Thread: f and g inverese of each other?

1. ## f and g inverese of each other?

am i doing this right or am i starting it the right way?

1st pic is the problem
2d and 3rd are what ive done

2. $\displaystyle f\left(\frac{6}{x-1}\right) =$

$\displaystyle \frac{6 + \frac{6}{x-1}}{\frac{6}{x-1}} =$

multiply numerator and denominator by $\displaystyle (x-1)$ to clear the fractions ...

$\displaystyle \frac{6(x-1) + 6}{6} =$

$\displaystyle \frac{6x - 6 + 6}{6} = x$

now ... you do $\displaystyle g[f(x)]$ , see if you get $\displaystyle x$ .

3. how do I clear the fractions at the bottom

6/[(6+x)/(x)] -1

so if I get x for g it makes then inverse?

4. Originally Posted by rj2001
how do I clear the fractions at the bottom

6/[(6+x)/(x)] -1
$\displaystyle \frac{6}{\frac{6+x}{x} - 1} \cdot \frac{x}{x}$

so if I get x for g it makes then inverse?
thought you already knew that ... if f[g(x)] = g[f(x)] = x, then f(x) and g(x) are inverses.

5. Could you just find the inverse of g(x) and show that it is f(x). It would be much quicker.

6. Originally Posted by whipflip15
Could you just find the inverse of g(x) and show that it is f(x). It would be much quicker.
Yes.

This is just another method to verify that two functions are inverses.