Find the Function G for Given F(x) = 1/x and (f/g)(x) = (x+1)/(x^2-x)

[jai6638]

Q) Given F(X) = 1/x and (f/g)(x) = (x+1)/(x^2-x). find the function g.

A: (x+1)/x(x-1) = (1/x) . (x+1)/(x-1)

Hence, the function G is (x+1)/(x-1)


Is this answer correct?

 

[HallsofIvy]

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

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

 

[jai6638]

So that would be (1/x) / (x+1)(x-1) = (x-1) / ( x^2+x)

Is that the correct answer?


thanks

 

[VietDao29]

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,

 

[HallsofIvy]

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

So $$\frac{1}{g(x)}= \frac{x+1}{x-1}$$.

Now do you see what g(x) is?

Once again, how do you divide by a fraction?

 

[jai6638]

damn... So does g(x) = (x-1)/(x+1) ?

By the way, at your first post, it reads F(x) = 1 / x. Do you really mean that or f(x) = 1 / x?

I mean, f(x)=1/x

 

[VietDao29]

Yup. That's correct.
Viet Dao,