1/f(x), why is there no point of discontinuity if the f(x) has a horizontal asymptote (1 Viewer)

[barbernator]

If we are told to graph 1/f(x), given f(x) in graphic form, why is there no point of discontinuity on the 1/f(x) graph at the point where there is a vertical asymptote in f(x)??

 

[Carrotsticks]

Re: 1/f(x), why is there no point of discontinuity if the f(x) has a horizontal asymp

If we are told to graph 1/f(x), given f(x) in graphic form, why is there no point of discontinuity on the 1/f(x) graph at the point where there is a horizontal asymptote in f(x)??

If it was a horizontal asymptote in f(x) for example y=2, then in the new curve 1/f(x), the horizontal asymptote becomes y=1/2 instead of y=2.

Points of discontinuity only occur if y=0 (since this makes the denominator 0), but a horizontal asymptote does not necessarily imply y=0 unless the x axis IS the actual asymptote itself.

 

[barbernator]

Re: 1/f(x), why is there no point of discontinuity if the f(x) has a horizontal asymp

If it was a horizontal asymptote in f(x) for example y=2, then in the new curve 1/f(x), the horizontal asymptote becomes y=1/2 instead of y=2.

Points of discontinuity only occur if y=0 (since this makes the denominator 0), but a horizontal asymptote does not necessarily imply y=0 unless the x axis IS the actual asymptote itself.

ahh sorry, i dont even know why i said horizontal asymptote, i mean vertical asymptote. Because today i assumed that if the function f(x) --> infinity, say x=2 is the vertical asymptote, there would be a point of discontinuity at x=2 on the function g(x) = 1/f(x) right? as there is no value for the original function at that point. But then my teacher said that at vertical asymptotes, the new function is just equal to zero, and im unsure if she was wrong or if there is a reason for it. Thanks!

 

[Carrotsticks]

Re: 1/f(x), why is there no point of discontinuity if the f(x) has a horizontal asymp

ahh sorry, i dont even know why i said horizontal asymptote, i mean vertical asymptote. Because today i assumed that if the function f(x) --> infinity, say x=2 is the vertical asymptote, there would be a point of discontinuity at x=2 on the function g(x) = 1/f(x) right? as there is no value for the original function at that point. But then my teacher said that at vertical asymptotes, the new function is just equal to zero, and im unsure if she was wrong or if there is a reason for it. Thanks!

Your teacher is correct.

Think about a vertical asymptote to be a VERY large y value ie: 1000000000.

We are doing 1/f(x) so we will be doing 1/100000000 etc *I think I am missing a zero but oh well*

What is 1 divded by a VERY large number? It approaches zero.

So a vertical asymptote becomes a 'root' with an open circle.

The reason why its an open circle is because the y co-ordinate never actually REACHES infinity, it just gets arbitrarily 'close to it' (technically not correct but should be okay for you atm).

Since the y coordinate of f(x) never actually reaches infinity, the y coordinate of 1/f(x) will never actually reach zero, hence an open circle.

 

[barbernator]

Re: 1/f(x), why is there no point of discontinuity if the f(x) has a horizontal asymp

my teacher said that there SHOULD be a value at that point, and i said there should be a point of discontinuity ahahah. But within the fitzpatrick solutions they never put in the points of discontinuity and im still unsure why.

 

[Carrotsticks]

Re: 1/f(x), why is there no point of discontinuity if the f(x) has a horizontal asymp

my teacher said that there SHOULD be a value at that point, and i said there should be a point of discontinuity ahahah. But within the fitzpatrick solutions they never put in the points of discontinuity and im still unsure why.

Oh sorry I misread what you had said.

Like I mentioned in my above post, it should NOT be a closed circle at that point.

You are correct, there is a point of discontinuity.

The limit EXISTS but the actual point doesn't. Does this make sense to you? Suppose x=a is where the vertical asymptote occurs. Here is a more precise mathematical definition: