- #1

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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?

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- Thread starter jai6638
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- #1

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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?

- #2

HallsofIvy

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

- #3

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

Is that the correct answer?

thanks

Is that the correct answer?

thanks

- #4

VietDao29

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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,

- #5

HallsofIvy

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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?

[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?

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- #6

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

I mean, f(x)=1/x

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

Last edited:

- #7

VietDao29

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Yup. That's correct.

Viet Dao,

Viet Dao,

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