Reciprocal Functions and Asymptotes

[DualDog321]

Here is the question in my textbook that I am struggling with:

I have a worked solutions book for it too, and it shows as follows:

I am unsure of where the new asymptotes like x=3, x=-2, and y=1/2 came from.
For anyone wondering, this is Chapter 27, question #84, on pg 846 in the Haese Math HL Third Edition textbook.
Thanks.

 

[MarkFL]

Hello, and welcome to FMH!

Look at the graph of the original function \(f(x)\). It has zeroes for \(x\in\{-2,3\}\). So, when this function is inverted, those zeros will become vertical asymptotes.

Because the original function has a horizontal asymptote at \(y=2\), the inverted function must have a horizontal asymptote at the reciprocal of \(2\), which is \(\dfrac{1}{2}\). The degree of the nummerator and denominator of both rational functions must be the same.

Likewise, the inverted function must have zeroes where the original has vertical asymptotes. And both functions will have the same sign wherever both are defined.

 

[DualDog321]

Ah, I see. Thank you, Mr. MarkFL. But how do we know/determine the end behavior for the inverted function?

 

[MarkFL]

Ah, I see. Thank you, Mr. MarkFL. But how do we know/determine the end behavior for the inverted function?

The original function has the horizontal asymptote of y = 2, and so the inverted function must have a horizontal asymptote at y = 1/2, since 1/2 is the multiplicative inverse of 2.

 

[JeffM]

Here is a different way to think about this. I do not have a good visual imagination. So I just think in small logical steps (well usually logical).

[MATH]-\ 9 \le x < - 3, f(x) \text { is increasing and positive} \implies \dfrac{1}{f(x)} \text { is positive and decreasing.} [/MATH]
Is that hard?

[MATH]x \rightarrow -\ 3 \implies f(x) \rightarrow +\ \infty \implies \dfrac{1}{f(x)} \rightarrow \dfrac{1}{+\ \infty} = 0.[/MATH]
In other words, the reciprocal approaches 0 from above because it is positive.

And f(- 3) does not exist. So what is 1/f(-3)?

Now think about [MATH]2 < x \le 9.[/MATH]
Then you deal with - 3 < x < x.

To summarize, you can break down logically what the graph means step by step. After all that is the purpose of a graph: to create a visual reminder of what is known mathematically. So you can always work backward from the graph to the mathematical meaning if you need to work with that meaning.

 

[DualDog321]

Here is a different way to think about this. I do not have a good visual imagination. So I just think in small logical steps (well usually logical).

[MATH]-\ 9 \le x < - 3, f(x) \text { is increasing and positive} \implies \dfrac{1}{f(x)} \text { is positive and decreasing.} [/MATH]
Is that hard?

[MATH]x \rightarrow -\ 3 \implies f(x) \rightarrow +\ \infty \implies \dfrac{1}{f(x)} \rightarrow \dfrac{1}{+\ \infty} = 0.[/MATH]
In other words, the reciprocal approaches 0 from above because it is positive.

And f(- 3) does not exist. So what is 1/f(-3)?

Now think about [MATH]2 < x \le 9.[/MATH]
Then you deal with - 3 < x < x.

To summarize, you can break down logically what the graph means step by step. After all that is the purpose of a graph: to create a visual reminder of what is known mathematically. So you can always work backward from the graph to the mathematical meaning if you need to work with that meaning.

Okay, that makes sense. Thanks for the help.