Function Inverses

[giddyupc]

Show that the functions are inverses.

f(x)= 3/(3-x) g(x)= (3x-x)/x

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When I solve f(g(x)) I get f(g(x))=x, but for some reason when I solve g(f(x)) I get g(f(x))= -x.

Could someone show me the steps to solve g(f(x)) please?
Thanks!

 

[giddyupc]

As you have written it, g(x) = (3x - x)/x = 2, which is not the inverse of f(x) = 3/(3 - x). This may explain why you are having trouble.

I'm sorry, I wrote g(x) incorrectly.

f(x)= 3/(3-x) and g(x)= (3x-3)/x

 

[lookagain]

Show that the functions are inverses.
\(\displaystyle f(x) \ = \ \frac{3}{3 - x} \ \ and \ \ g(x) \ = \ \frac{3x - 3}{x}\)

When I solve f(g(x)) I get f(g(x))=x, but for some reason when I solve g(f(x)) I get g(f(x))= -x.
Could someone show me the steps to solve g(f(x)) please? Thanks!

\(\displaystyle g(x) \ = (3x - 3)\bigg(\frac{1}{x}\bigg)\)


\(\displaystyle g(f(x)) \ = \)


\(\displaystyle \ \bigg[(3)\bigg(\frac{3}{3 - x}\bigg) - 3\bigg]\bigg(\frac{3 - x}{3}\bigg) ** \ =\)


\(\displaystyle \bigg(\frac{3}{1}\bigg)\bigg(\frac{3}{3 - x}\bigg)\bigg(\frac{3 - x}{3}\bigg) \ - \ \bigg(\frac{3}{1}\bigg)\bigg(\frac{3 - x}{3}\bigg)\)


\(\displaystyle 3 - (3 - x) \ = \)


\(\displaystyle 3 - 3 + x \ = \)


\(\displaystyle x\)




\(\displaystyle \text{** Note: When 3/(3 - x) is substituted into 1/x, the result is (3 - x)/3.}\)