inverting a function, quick question (1 Viewer)

sadpwner

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Will the function exist at the vertical asymptote after I invert it? I thought it did, but I was doing revision yesterday and my teacher circled the coordinates of the vertical asymptotes.
 

integral95

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if i recall correctly, the vertical asymptote would become a horizontal asymptote.

e.g if the original function as x = 1 as the asymptote, then the inverse function would have an asymptote at y = 1
 

funnytomato

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Will the function exist at the vertical asymptote after I invert it? I thought it did, but I was doing revision yesterday and my teacher circled the coordinates of the vertical asymptotes.
if your teacher "circled the coordinate", I think( just taking a wild guess) what you mean is taking the reciprocal like , not the inverse

For example, look at this question from 2014 HSC. In part (ii) you are gonna have an open circle because the function is not defined at , neither is its reciprocal




P.S. I had similar doubts as well when I learned it, because intuitively it is approaching that point. However, the domain is restricted so we can only have an open circle.

Note for the function , you do not treat it as the reciprocal of especially when it comes to graphing. That is why you don't see the open circles for these graphs.
 
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