Graph of a function and its reciprocal

[ScienceJen]

Hi there,

I'm taking calculus because it is mandatory for my Bachelor of Science and I'm learning through an online university. Math has never been my strong suit and there are many concepts I'm having trouble wrapping my head around.
Can someone please explain to me why the reciprocal graph looks the way it does? I especially don't understand the 1/b line - isn't the shape of that the same as graphing the function of a square root?



Thank you,

Jen

 

[Dr.Peterson]

I'm taking calculus because it is mandatory for my Bachelor of Science and I'm learning through an online university. Math has never been my strong suit and there are many concepts I'm having trouble wrapping my head around.
Can someone please explain to me why the reciprocal graph looks the way it does? I especially don't understand the 1/b line - isn't the shape of that the same as graphing the function of a square root?

First, I need to point out that there is no "1/b line". The label 1/b refers to the point (0, 1/b), which corresponds to the point (0, b) in the original graph. Similarly, a and c refer to x-coordinates, which in the original are locations of x-intercepts, and in the reciprocal graph are locations of vertical asymptotes.

Second, it asks a question, "Do you agree with them?"; I suppose they expect you to say yes, after examining them and thinking about what they say. I don't really agree, because they are rather poorly drawn! The left and right parts are not drawn as functions, because the curve away from the asymptotes; and the middle part appears to drop a little at the start, which it should never do. So if it happens to be really intended as a question, you might want to say no!

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Here is a more accurate graph of such a function (green is the original, g; red is the reciprocal, 1/g):

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But other than such quibbles, you should see that the graph makes sense. What you need to do in order to see it (and can do better on my accurate graph) is to pick some value of x and observe that y on my red curve is the reciprocal of y on my green curve.

More generally, observe that where g(x) = 0, 1/g(x) is undefined, because you can't divide by zero; in particular, in terms of limits, for example, as x approaches 2 from the left, the green line has positive values approaching zero, while the red curve has positive values approaching infinity. Wherever the green line is decreasing, the red line is increasing; wherever the green line is moving toward zero, the red curve is moving away from zero.

If that doesn't help, we'll want to hear more from you about your specific confusions. In particular, can you say more about the square root part?

 

[ScienceJen]

First, I need to point out that there is no "1/b line". The label 1/b refers to the point (0, 1/b), which corresponds to the point (0, b) in the original graph. Similarly, a and c refer to x-coordinates, which in the original are locations of x-intercepts, and in the reciprocal graph are locations of vertical asymptotes.

Second, it asks a question, "Do you agree with them?"; I suppose they expect you to say yes, after examining them and thinking about what they say. I don't really agree, because they are rather poorly drawn! The left and right parts are not drawn as functions, because the curve away from the asymptotes; and the middle part appears to drop a little at the start, which it should never do. So if it happens to be really intended as a question, you might want to say no!

View attachment 31760​

Here is a more accurate graph of such a function (green is the original, g; red is the reciprocal, 1/g):

View attachment 31761​

But other than such quibbles, you should see that the graph makes sense. What you need to do in order to see it (and can do better on my accurate graph) is to pick some value of x and observe that y on my red curve is the reciprocal of y on my green curve.

More generally, observe that where g(x) = 0, 1/g(x) is undefined, because you can't divide by zero; in particular, in terms of limits, for example, as x approaches 2 from the left, the green line has positive values approaching zero, while the red curve has positive values approaching infinity. Wherever the green line is decreasing, the red line is increasing; wherever the green line is moving toward zero, the red curve is moving away from zero.

If that doesn't help, we'll want to hear more from you about your specific confusions. In particular, can you say more about the square root part?

Good morning Dr. Peterson,

Thank you so much for your quick and detailed response! I really appreciate it.

That certainly makes more sense now - especially after explaining "wherever the green line is decreasing, the red line is increasing; wherever the green line is moving toward zero, the red curve is moving away from zero."

However, there is something else: I don't quite understand how the shape of the reciprocal lines are determined and how many lines there should be. With regards to the square root part that I mentioned, I am referring to the basic shape of a graph for the function sqrt x. To me, it looks like the line coming from the point 1/b, but starts at zero and has positive values as it moves along the x axis.
Thank you for your time,

Jennifer

 

[Dr.Peterson]

With regards to the square root part that I mentioned, I am referring to the basic shape of a graph for the function sqrt x. To me, it looks like the line coming from the point 1/b, but starts at zero and has positive values as it moves along the x axis.

Really, the middle part looks more like $x^2$ than $\sqrt{x}$! But either one is irrelevant.

I don't quite understand how the shape of the reciprocal lines are determined and how many lines there should be.

You will probably benefit from making a graph of 1/x, or something else like 2/(x-2), by hand, so you can get an experiential feel for such functions. Make a table of values.

 

[ScienceJen]

Really, the middle part looks more like $x^2$ than $\sqrt{x}$! But either one is irrelevant.

You will probably benefit from making a graph of 1/x, or something else like 2/(x-2), by hand, so you can get an experiential feel for such functions. Make a table of

I don't know how to do that. I'm taking calculus after graduating high school in 1999 and I just told stats last year - going for my BSc. Math has never been a strong point.

 

[Dr.Peterson]

I don't know how to do that. I'm taking calculus after graduating high school in 1999 and I just told stats last year - going for my BSc. Math has never been a strong point.

All I'm suggesting you do is to make a table of values and plot points. For y = 1/x, you might take values of x as, say, 1/4, 1/2, 3/4, 1, 2, 3, 4. The corresponding values of y are 4, 2, 1 1/3, 1, 1/2, 1/3, 1/4. So you'd plot the points (1/4, 4), (1/2, 2), (3/4, 1 1/3), (1, 1), (2, 1/2), (3, 1/3), (4, 1/4). You should also do the same with negative numbers. Then connect points with a curve, and compare your results with what you see when you enter y=1/x on desmos.com.

This should demystify the graphing process, getting you in closer touch with what it means for a function to increase or decrease.

You really need to have this algebra knowledge already before taking calculus (though I often say that calculus is where many students finally really learn algebra, because they're forced to use it as a tool). If you have access to tutoring at your school, you should look into it; that's what I do, and this semester I've watched one student in a similar situation to yours grow from questioning whether she'd pass at all (this is precalculus) to getting an A currently. She just needed help remembering what she had forgotten, just in time to keep up with the class.

 

[ScienceJen]

All I'm suggesting you do is to make a table of values and plot points. For y = 1/x, you might take values of x as, say, 1/4, 1/2, 3/4, 1, 2, 3, 4. The corresponding values of y are 4, 2, 1 1/3, 1, 1/2, 1/3, 1/4. So you'd plot the points (1/4, 4), (1/2, 2), (3/4, 1 1/3), (1, 1), (2, 1/2), (3, 1/3), (4, 1/4). You should also do the same with negative numbers. Then connect points with a curve, and compare your results with what you see when you enter y=1/x on desmos.com.

This should demystify the graphing process, getting you in closer touch with what it means for a function to increase or decrease.

You really need to have this algebra knowledge already before taking calculus (though I often say that calculus is where many students finally really learn algebra, because they're forced to use it as a tool). If you have access to tutoring at your school, you should look into it; that's what I do, and this semester I've watched one student in a similar situation to yours grow from questioning whether she'd pass at all (this is precalculus) to getting an A currently. She just needed help remembering what she had forgotten, just in time to keep up with the class.

Thank you so much for taking the time to explain that to me. Unfortunately, I'm taking calculus through an online university, as I did with stats and my "tutor" (the person assigned to "help" me) has informed me that he has 160 student and that I should "'dig my heels in more."' I'm trying really hard. Thank goodness for people like you on this forum, Khan Academy, and YouTube videos.