Question about local linearization?

[imwishing73]

I just discovered this site...I feel like I'll be getting to know you all very quickly!

Anyways, for all values of x in the interval containing the point of tangency, if the local linearization of the function f is always less than or equal to the values of the function, then which of the following is true?
A. f'(x)<0
B. f''(x)<0
C. f'(x)>0
D. f''(x)=0
E. f''(x)>0

I have no idea where to even begin with this problem, so a push in the right direction might be all I need. What exactly does "interval containing the point of tangency" refer to, in more dumbed-down language?
Thank you!!

 

[Deleted member 4993]

I just discovered this site...I feel like I'll be getting to know you all very quickly!

Anyways, for all values of x in the interval containing the point of tangency, if the local linearization of the function f is always less than or equal to the values of the function, then which of the following is true?
A. f'(x)<0
B. f''(x)<0
C. f'(x)>0
D. f''(x)=0
E. f''(x)>0

I have no idea where to even begin with this problem, so a push in the right direction might be all I need. What exactly does "interval containing the point of tangency" refer to, in more dumbed-down language?
Thank you!!

the point of tangency - did they refer to a tangent before posing this problem?

 

[imwishing73]

No reference...that's all the information that was given. Does it even matter, though? The answer is just an inequality.

 

[JeffM]

No reference...that's all the information that was given. Does it even matter, though? The answer is just an inequality.

One nice thing about multiple choice questions is that (unless they include the "none of the above" trick) you know one of the proposed answers is correct. And unless there is choice that reads like "A and C above," you also know that only one of proposed answers is correct. That is always a HUGE clue.

It is very hard to give help on this question without just giving out the answer, in which case you learn nothing from the exercise.

Sketching some diagrams will help. Hint: A differentiable curve is "simple" if you look at a small enough interval.

Considering two conceptual questions will help in drawing these diagrams and interpreting their meaning.

What two relationships exist between the tangent to a curve and the curve itself at the point of tangency?

Also, what is the relationship between between the first and second derivative? To put it a different way, what does it imply if the second derivative is negative? What if the second derivative is positive?

 

[imwishing73]

One nice thing about multiple choice questions is that (unless they include the "none of the above" trick) you know one of the proposed answers is correct. And unless there is choice that reads like "A and C above," you also know that only one of proposed answers is correct. That is always a HUGE clue.

It is very hard to give help on this question without just giving out the answer, in which case you learn nothing from the exercise.

Sketching some diagrams will help. Hint: A differentiable curve is "simple" if you look at a small enough interval.

Considering two conceptual questions will help in drawing these diagrams and interpreting their meaning.

What two relationships exist between the tangent to a curve and the curve itself at the point of tangency?

Also, what is the relationship between between the first and second derivative? To put it a different way, what does it imply if the second derivative is negative? What if the second derivative is positive?

Would the answer be f'(x)<0? If I sketch it out, at a local level, the y-value is always greater than the local linearization, if I'm thinking about this correctly. As for the questions you asked about the relationship between the first and second derivative, I honestly have no idea. However, I would think that if the second derivative was negative, then the first derivative would have to be decreasing, which means that the function is negative? Does that make ANY sense?

 

[JeffM]

Would the answer be f'(x)<0? If I sketch it out, at a local level, the y-value is always greater than the local linearization, if I'm thinking about this correctly. As for the questions you asked about the relationship between the first and second derivative, I honestly have no idea. However, I would think that if the second derivative was negative, then the first derivative would have to be decreasing, which means that the function is negative? Does that make ANY sense?

OK you are thinking along promising lines.

If you sketch the curve (a simple curve), the premise of the question is that the curve will always be higher than the line of tangency except at the point of tangency, where the the line and the curve will be equal.

Now if you draw a curve that looks like \(\displaystyle \cap\), the tangent will be higher than the curve no matter where you place the point of tangency.

So the curve being considered looks like \(\displaystyle \cup\).

The slope of the tangent line equals the slope of the curve at the point of tangency, and the slope of the curve equals the first derivative, right?

So, considering ANY point on the relevant curve, what if anything can you say about the slope of the tangent and therefore about the first derivative?

 

[imwishing73]

OK you are thinking along promising lines.

If you sketch the curve (a simple curve), the premise of the question is that the curve will always be higher than the line of tangency except at the point of tangency, where the the line and the curve will be equal.

Now if you draw a curve that looks like \(\displaystyle \cap\), the tangent will be higher than the curve no matter where you place the point of tangency.

So the curve being considered looks like \(\displaystyle \cup\).

The slope of the tangent line equals the slope of the curve at the point of tangency, and the slope of the curve equals the first derivative, right?

So, considering ANY point on the relevant curve, what if anything can you say about the slope of the tangent and therefore about the first derivative?

Well...the slope is negative when x<0 and positive when x>0, right?
But the first derivative would be positive either way...did the lightbulb just go on above my head?

 

[JeffM]

Well...the slope is negative when x<0 and positive when x>0, right?
But the first derivative would be positive either way...did the lightbulb just go on above my head?

No. You are in the dark.

Nothing in the problem indicates that the interval under consideration requires that x > 0 or x < 0. We have not a clue about the sign of x in the interval.

All that the problem specifies is that there exists a point a and a small interval around it such that for every x in the interval not equal to a
f(x) - t(x) > 0, where t(x) is the equation describing the line of tangency at point a. (At least this is how I imagine the problem is approximately worded. You did not give the exact wording.)

By definition the slope of the tangent line equals the first derivative of the curve at the point of tangency. Now is there anything in what is specified that says or implies something about the sign of the tangent at point a.

 

[imwishing73]

No. You are in the dark.

Nothing in the problem indicates that the interval under consideration requires that x > 0 or x < 0. We have not a clue about the sign of x in the interval.

All that the problem specifies is that there exists a point a and a small interval around it such that for every x in the interval not equal to a
f(x) - t(x) > 0, where t(x) is the equation describing the line of tangency at point a. (At least this is how I imagine the problem is approximately worded. You did not give the exact wording.)

By definition the slope of the tangent line equals the first derivative of the curve at the point of tangency. Now is there anything in what is specified that says or implies something about the sign of the tangent at point a.

I gave the problem word-for-word. Any information that I was given was passed on. So no, but I see my mistake. In my mind, I was envisioning y=x^2 just to give the problem some sort of visual aid. But if we're looking at a curve that looks like \(\displaystyle \cup\), the slope is decreasing on the left and increasing on the right. THAT'S the part that confuses me. How do you get one answer?

 

[JeffM]

I gave the problem word-for-word. Any information that I was given was passed on. So no, but I see my mistake. In my mind, I was envisioning y=x^2 just to give the problem some sort of visual aid. But if we're looking at a curve that looks like \(\displaystyle \cup\), the slope is decreasing on the left and increasing on the right. THAT'S the part that confuses me. How do you get one answer?

OK. You are back on track.

I am not crazy about the wording of this problem so I am going to proceed on what I suspect is the intention behind it.

You are absolutely right that for your U-shaped curve the slope of the tangent is decreasing at any point to the left of the bottom of the U and is increasing at any point to the right of the bottom of the U. So is the sign of the slope of the tangent line determined by the specifications of the problem or not? So what does that imply about the sign of the first derivative? So can you eliminate any of the possible answers you were given?

By the way, the curve is not necessarily U-shaped. It COULD look like the left-most part or like the right-most part of the U. But it can be U-shaped. That is, the problem does not put much in the way of definition on the curve so you must not get too caught up in a very specific visualization.

 

[imwishing73]

OK. You are back on track.

I am not crazy about the wording of this problem so I am going to proceed on what I suspect is the intention behind it.

You are absolutely right that for your U-shaped curve the slope of the tangent is decreasing at any point to the left of the bottom of the U and is increasing at any point to the right of the bottom of the U. So is the sign of the slope of the tangent line determined by the specifications of the problem or not? So what does that imply about the sign of the first derivative? So can you eliminate any of the possible answers you were given?

By the way, the curve is not necessarily U-shaped. It COULD look like the left-most part or like the right-most part of the U. But it can be U-shaped. That is, the problem does not put much in the way of definition on the curve so you must not get too caught up in a very specific visualization.

Okay, so either way, the sign of the slope has to be determined by the specifications. So it could potentially be negative, eliminating C, D, and E.

No matter what, we're always looking at less than or equal local linearization. But the velocity could still be increasing, while the acceleration has to be decreasing, so f"(x) has to be negative?

 

[Deleted member 4993]

I think you have missed the class on concave/convex nature of curves and the connection of radius of curvature to second derivative of the function.

Think a bit -

if the linearized approximation is always less than the actual value within a given domain - is the curve concave or convex? What can you deduce about its second derivative?

if the linearized approximation is always more than the actual value within a given domain - is the curve concave or convex? What can you deduce about its second derivative?

I think - the reference to tangency is to assure that the curve is continuous within the region (sometimes the examiners get too 'smart' for their own good) .

 

[imwishing73]

I think you have missed the class on concave/convex nature of curves and the connection of radius of curvature to second derivative of the function.

Think a bit -

if the linearized approximation is always less than the actual value within a given domain - is the curve concave or convex? What can you deduce about its second derivative?

if the linearized approximation is always more than the actual value within a given domain - is the curve concave or convex? What can you deduce about its second derivative?

I think - the reference to tangency is to assure that the curve is continuous within the region (sometimes the examiners get too 'smart' for their own good) .

We never really elaborated on the second derivative in class, other than the fact that it is the derivative of the first derivative, so that is definitely my problem. I am also still struggling immensely to picture the graph of a derivative for any given function. If the approximation is less, then the curve is concave, and if the approximation is more than the value, then the curve is convex. I can't quite grasp what that means about the first derivative, let alone the second derivative, other than the fact that the first derivative would be zero at the turning point and increasing and decreasing at either side.

 

[JeffM]

Okay, so either way, the sign of the slope of the tangent line at a point has to be determined by the specifications. So the slope of the tangent line at the point of tangency (and therefore the sign of the first derivative) could potentially be negative, or positive, or zero. The problem does not give enough information to tell.

eliminating C, D, and E. Huh? The only thing we have discussed so far is the FIRST derivative, and answers D and E (as well as B) concern the second derivative. Only A and C involve the first derivative. So we can eliminate A and C because they specify the sign of the first derivative, and we know that we have insufficient information to make such a specification.

No matter what, we're always looking at less than or equal local linearization. But the velocity could still be increasing, while the acceleration has to be decreasing, so f"(x) has to be negative? No.

As I said before, we do not necessarily have a U. We could have the left-hand side of the U, or the right-hand side, or the very bottom. We have to consider all three cases.

Let's take the simplest case: say that the point of tangency, point a, is at the very bottom of the U. Now what is the slope of the curve to the left of a? What is the slope of the curve to the right of a. So what is the slope at a? As x increases, what happens to the slope of the curve? So what is happening to the first derivative? Now the second derivative is the slope of the first derivative. So can we tell what is the sign of the second derivative?

Edit: The first derivative of function f(x) is itself function that gives the slope of the curve of f(x) at every point. The slope or first derivative gives some important information about the function f(x) behaves. But the first derivative does not tell us everything.

But the first derivative is itself a function and (if it is differentiable) has a slope, namely the second derivative of f(x), which tells us how the slope of the function f(x) is changing. But this gives us even more information about how the function is changing. I know: it all seems a bit abstract, but you can think of each successive derivative as giving ever more refined information about how the underlying function f(x) behaves.

Edit 2: Take some functions. Graph them. Calculate their first derivatives and graph them. The first derivative is just a function.

 

[imwishing73]

As I said before, we do not necessarily have a U. We could have the left-hand side of the U, or the right-hand side, or the very bottom. We have to consider all three cases.

Let's take the simplest case: say that the point of tangency, point a, is at the very bottom of the U. Now what is the slope of the curve to the left of a? What is the slope of the curve to the right of a. So what is the slope at a? As x increases, what happens to the slope of the curve? So what is happening to the first derivative? Now the second derivative is the slope of the first derivative. So can we tell what is the slope of the second derivative?

The slope to the left is negative, to the right is positive, and at a, is zero. As x increases, the slope decreases, then becomes zero, then increases. That means the first derivative is a straight line, so the second derivative is a constant, and f"x=0 is the only option that's a constant...?

 

[JeffM]

The slope to the left is negative, YES to the right is positive, YES and at a, is zero. YES As x increases, the slope decreases, then becomes zero, then increases. No, not at all. As approaches the bottom from the left, the slope becomes LESS negative, becomes 0 at the bottom, and then becomes more positive. That means the first derivative is a straight line, No, it means that the first derivative is increasing in that interval. It may or may not be a straight line. so the second derivative is a constant, Not necessarily. Only if the first derivative is a straight line. and f"x=0 is the only option that's a constant...? So do you see why this is not right.

You are getting closer.

If the function f'(x) that describes the slope of our function f(x) is increasing near point a, then what is the sign of the slope of f'(x).

 

[imwishing73]

You are getting closer.

If the function f'(x) that describes the slope of our function f(x) is increasing near point a, then what is the sign of the slope of f'(x).

Well, I only have two more options left. Let's hope I can get it on this try. I get what you're saying. So the first derivative is increasing, which means the second derivative is increasing, which means it must be positive., so f''(x)>0.

 

[JeffM]

Well, I only have two more options left. Let's hope I can get it on this try. I get what you're saying. So the first derivative is increasing, which means the second derivative is increasing, which means it must be positive., so f''(x)>0.

Well you now have the right answer for ONE of the three cases, but there is a slight slip in your reasoning.

If the first derivative is increasing at a point, that implies that the second derivative is NOT NEGATIVE. It does not imply anything about whether the second derivative is increasing or decreasing at that point. As pointed out, the wording of the problem is not ideal, but I think that the intent is for you to conclude that the second derivative is positive.

You are not done yet because we took the easiest case, where the point of tangency was at the bottom of the U. But as I have explained, the curve may look like just the left-hand side of a U without a bottom in the relevant interval, or it may look like the right-hand side of a U without a bottom in the relevant interval. So you really should analyze these two cases.

 

[HallsofIvy]

If the graph of a function lies above its tangent line at a point, then the function is concave upward. What does that tell you about the sign of the second derivative?