f and g inverese of each other? - Math Help Forum

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#1

$Old$ November 21st, 2008, 04:01 PM

rj2001 $rj2001 is offline$

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$Default$ 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

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#2

$Old$ November 21st, 2008, 04:15 PM

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

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

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

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

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

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

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#3

$Old$ November 21st, 2008, 06:30 PM

rj2001 $rj2001 is offline$

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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?

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$Old$ November 21st, 2008, 06:39 PM

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Originally Posted by rj2001 $View Post$

how do I clear the fractions at the bottom

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

$\frac{6}{\frac{6+x}{x} - 1} \cdot \frac{x}{x}$

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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.

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$Old$ November 21st, 2008, 07:02 PM

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

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$Old$ November 21st, 2008, 08:41 PM

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Originally Posted by whipflip15 $View Post$

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.

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