Creating f graph based on f' graph

[deleted_user]

I am not sure if I have the right answers, I'm confused as to how f will be when f'(x) approaches a vertical asymptote. Here is what i have so far. POI is point of inflection.

 

[Dr.Peterson]

I am not sure if I have the right answers, I'm confused as to how f will be when f'(x) approaches a vertical asymptote. Here is what i have so far. POI is point of inflection.
View attachment 32762

How can a relative max or min be a point of inflection? And why do you have a POI at 37/10?

It would be more helpful to see your attempt at a graph, which we could check out more easily. In your table, I would expect to see mentions of concavity, and perhaps of sharp turns.

But what you say for x=3 is correct, just perhaps not as complete as you wish. Please either tell us more about what you think happens there, or show us your attempt at a graph. Either way, we'll have a better idea of how you are confused.

 

[Steven G]

I am not sure if I have the right answers, I'm confused as to how f will be when f'(x) approaches a vertical asymptote. Here is what i have so far. POI is point of inflection.
View attachment 32762

From -oo to x=-2, the derivative goes from oo to 1. That is on f(x), the slope goes from oo (ie vertical) to 1. Now draw that from -oo to 1.
From x=-2 to 3, the derivative goes from -2 to oo. That is on f(x), the slope goes from -2 to oo (vertical). Draw this
From x=3 to x=oo, the derivative goes from oo to 0. That is on f(x), the slope goes from oo (ie vertical) to 0. Draw this.

You do not need a table to do such a problem.

 

[JeffM]

The point I would stress is that there is no hint that this problem has a unique answer. It appears that you are permitted to choose plausible functions to work with. For example, the rightmost segment resembles a hyperbola so assume it is a hyperbola. What might the function describing that hyperbola be? From what kind of functions would you get such a derivative?

 

[Dr.Peterson]

The point I would stress is that there is no hint that this problem has a unique answer.

Absolutely. In particular, you can start at any point you want, and pay attention only to the general shape. It's only a possible sketch.

It appears that you are permitted to choose plausible functions to work with. For example, the rightmost segment resembles a hyperbola so assume it is a hyperbola. What might the function describing that hyperbola be? From what kind of functions would you get such a derivative?

I'd go beyond that: We don't even need to imagine (the pieces of) the function having equations at all, which could lead to worrying about integrating them. I'd just sketch a curve that is, say, increasing and concave down in one region, culminating in slope of 1, and so on. Just as Steven G suggests: the table of information is overkill.

 

[JeffM]

I'd go beyond that: We don't even need to imagine (the pieces of) the function having equations at all, which could lead to worrying about integrating them.

True, and highly pertinent if the student has not been introduced to integration.

But the rightmost and leftmost portions of the curve become quite tractable with integration. I jumped to the conclusion that the student was familiar with integration. (My wife claims I get very little exercise other than jumping to conclusions. It is not true. I lift one or two martinis daily.)

But this problem is a great introduction to integration.

 

[Steven G]

You don't need to have studied anything about integration to know how to solve basic antiderivates.
One of the pieces of f'(x) looks like 1/x to me. A student who never studied integrals should now that if f'(x)= 1/x, then f(x) = ln x.
In my old age, I am leaning towards expecting more from students. I have clearly seen that when you raise the bar (of standards), that students will do better than if the bar was lowered.

 

[deleted_user]

How can a relative max or min be a point of inflection? And why do you have a POI at 37/10?

It would be more helpful to see your attempt at a graph, which we could check out more easily. In your table, I would expect to see mentions of concavity, and perhaps of sharp turns.

But what you say for x=3 is correct, just perhaps not as complete as you wish. Please either tell us more about what you think happens there, or show us your attempt at a graph. Either way, we'll have a better idea of how you are confused.

i see, an point of inflection would be when concavity changes, so they wouldn't be a relative maximum and minimum.
yes, i was confused about 37/10, but i realized it was not changing concavity its just further going down to positive infinity.
i forgot to place the concavities, but negative f''(x) would be concave downs and concave up for positive f''(x).
i am not sure about sharp turns, but would it be where the vertical asymptote of 3 is? when the value of x approaches 3, the value of y increases to positive infinity. but it would never reach 3 and be undefined there.

and here is what i think the graph would be.

 

[JeffM]

You don't need to have studied anything about integration to know how to solve basic antiderivates.
One of the pieces of f'(x) looks like 1/x to me. A student who never studied integrals should now that if f'(x)= 1/x.

Sort of my point. The idea of the inverse of a derivative is inherent in the problem. Now if the student has not been introduced to anti-derivatives, this problem may spark an “ah-ha moment.” If the student has been introduced to anti-derivatives, this problem should spark an “ah-ha“ moment. But Dr. Peterson was right to correct me. You do not need anti-derivatives to address this problem. More basic tools suffice. But if you own a bulldozer, you tend to forget about shovels.

 

[deleted_user]

From -oo to x=-2, the derivative goes from oo to 1. That is on f(x), the slope goes from oo (ie vertical) to 1. Now draw that from -oo to 1.
From x=-2 to 3, the derivative goes from -2 to oo. That is on f(x), the slope goes from -2 to oo (vertical). Draw this
From x=3 to x=oo, the derivative goes from oo to 0. That is on f(x), the slope goes from oo (ie vertical) to 0. Draw this.

You do not need a table to do such a problem.

our teacher does tables, so i assumed it would be the right way to go.
i adjusted the graph such that (3,+inf) does not get to 0 just like the f'(x) graph.
the y at -2 below would be at y -2 too?

 

[Dr.Peterson]

i forgot to place the concavities, but negative f''(x) would be concave downs and concave up for positive f''(x).
i am not sure about sharp turns, but would it be where the vertical asymptote of 3 is? when the value of x approaches 3, the value of y increases to positive infinity. but it would never reach 3 and be undefined there.

and here is what i think the graph would be.

i adjusted the graph such that (3,+inf) does not get to 0 just like the f'(x) graph.
the y at -2 below would be at y -2 too?

First, you missed the fact that f is supposed to be continuous everywhere! Remove those discontinuities. In doing so, you may discover what I mean about sharp turns. (This is also a reason that the idea of thinking of the $(3,\infty)$ portion as a hyperbola can be counterproductive.)

Also, it doesn't matter what the actual values of y are in your graph; the whole thing could be shifted up or down without affecting the derivative. (That's a point you'll learn when you get to integration and the antiderivative.)

Finally, a much more minor point, your relative minimum doesn't look like it's at the right place.

I'm not sure what you mean by "y at -2 below would be at y -2". Do you mean the graph to the left of x=-2 should be a constant, y=-2 ??

 

[deleted_user]

First, you missed the fact that f is supposed to be continuous everywhere! Remove those discontinuities. In doing so, you may discover what I mean about sharp turns. (This is also a reason that the idea of thinking of the (3,\infty)(3,∞) portion as a hyperbola can be counterproductive.)

do you mean like this? oh the right part should not go beyond 0

Also, it doesn't matter what the actual values of y are in your graph; the whole thing could be shifted up or down without affecting the derivative. (That's a point you'll learn when you get to integration and the antiderivative.)

ooh i see, i believe we'll be learning this in the next school year!

I'm not sure what you mean by "y at -2 below would be at y -2". Do you mean the graph to the left of x=-2 should be a constant, y=-2 ??

i meant will the open points at the graph of f'(x) be the same placement as at the f(x) graph?

Finally, a much more minor point, your relative minimum doesn't look like it's at the right place.

since it is an x-intercept at the f'(x) graph, its a stationary point, so it would be the same placement in f(x)?

 

[Steven G]

continuous everywhere

 

[deleted_user]

continuous everywhere

i think i get it!

 

[Steven G]

As x approaches -2 from the left, the slope should be 1. As x approaches -2 from the right, the slope should be -2. In your graph the slope from the left and right of -2 seems to be approaching 0. Try again.

 

[Steven G]

Also, at x=13/5, the y values does not have to be 0. It can be 0, but again it is not necessary.

 

[Steven G]

Do you see how you could have gotten the same graph (without the error I already pointed out) just with the information I gave you in my previous post?
From -oo to x=-2, the derivative goes from oo to 1. That is on f(x), the slope goes from oo (ie vertical) to 1. Now draw that from -oo to 1.
From x=-2 to 3, the derivative goes from -2 to oo. That is on f(x), the slope goes from -2 to oo (vertical). Draw this
From x=3 to x=oo, the derivative goes from oo to 0. That is on f(x), the slope goes from oo (ie vertical) to 0

 

[deleted_user]

Thank you everyone who helped me! I'll end it here.