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if f(x)= 1/x+2, what is f^-1(x)?
f(x)=y=1/x+2 you need to solve for x with these inverse problems which just means rearranging your equation to isolate x on one side of the equations since y=1/x+2 your next step would be to subtract the 2 y-2=1/x your next step would be to multiply by x x(y-2)=1 then divide by y-2 x=1/(y-2)
i got \[f^-1(x)=\frac{ 1 }{ x-2 }\]

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y should be on the bottom 1/y-2
\[\large f(x)=\frac{1}{x+2}\] \[\large g(x)=\frac{1}{x}+2\] Which one is the correct format, f or g? :)
f?
\[\large \text{Let}\quad f(x)=y, \qquad \rightarrow \qquad y=\frac{1}{x+2}\]Taking the inverse gives us,\[\large x=\frac{1}{y+2}\]Multiply both sides by (y+2), giving us,\[\large x(y+2)=1\]Divide both sides by x,\[\large (y+2)=\frac{1}{x}\]Subtract 2 from each side,\[\large y=\frac{1}{x}-2\]Rewrite with inverse notation,\[\large f^{-1}(x)=\frac{1}{x}-2\]
I came up with something slightly different katie :D lemme know if those steps make sense.
i see thank you(:

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