Function Homework Help

Almost! You have shown that [itex]\frac{x+1}{x^2- x}= \frac{x+1}{x(x-1)}= \frac{1}{x}\left(\frac{x+1}{x-1}\right)= f(x)\left(\frac{x+1}{x-1}\right)[/itex] correctly.

But you want f/g, not f*g! How do you divide by a fraction?

 

Nope, that's still wrong. You can recheck your calculation.
g(x) = f(x) / ((f / g)(x))
By the way, at your first post, it reads F(x) = 1 / x. Do you really mean that or f(x) = 1 / x?
Viet Dao,

 

What I wrote before was
[tex]\frac{x+1}{x^2- x}= \frac{x+1}{x(x-1)}= \frac{1}{x}\left(\frac{x+1}{x-1}\right)= f(x)\left(\frac{x+1}{x-1}\right)= \frac{f(x)}{g(x)}[/tex]

So [tex]\frac{1}{g(x)}= \frac{x+1}{x-1}[/tex].

Now do you see what g(x) is?

Once again, how do you divide by a fraction?

 

Yup. That's correct.
Viet Dao,