Find the inverse of f(x)=1 over x-2. I know you swap x and y to find your inverse but i think I'm going about it wrong.
show what you have so far
I'm getting f^-1(x)=x+2
I'll get you started \[\large f(x) = \frac{1}{x-2}\] \[\large y = \frac{1}{x-2}\] \[\large x = \frac{1}{y-2}\] \[\large x(y-2) = 1\] The goal is to solve for y.
Sooo if i did my math correctly, is should be y=1/x+2?
let's finish up and find out \[\large f(x) = \frac{1}{x-2}\] \[\large y = \frac{1}{x-2}\] \[\large x = \frac{1}{y-2}\] \[\large x(y-2) = 1\] \[\large xy-2x = 1\] \[\large xy = 1+2x\] \[\large y = \frac{1+2x}{x}\] \[\large f^{-1}(x) = \frac{1+2x}{x}\]
You can rewrite the answer like this \[\large f^{-1}(x) = \frac{1+2x}{x}\] \[\large f^{-1}(x) = \frac{1}{x}+\frac{2x}{x}\] \[\large f^{-1}(x) = \frac{1}{x}+2\]
so it looks like you got \[\large f^{-1}(x) = \frac{1}{x}+2\]
Yay!!! :) your awesome ! Is it okay if you start me off on a similar one ?
sure, did you have one?
Yeah, its find the inverse of f(x)=2x-4.
When i first tried it i got y=x+4 over 2. Buut, im not so sure on it anymore.
\[\large f(x)=2x-4\] \[\large y=2x-4\] \[\large x=2y-4\] \[\large x+4=2y\] \[\large 2y=x+4\] \[\large y=\frac{x+4}{2}\] \[\large f^{-1}(x)=\frac{x+4}{2}\] So you nailed it.
Thank you soo much !! It means alot :)
sure thing
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