How does one find the maximum value of f given the graph of f'

[://Justice]

Homework Statement

The specific problem can be found here: http://www.cbsd.org/sites/teachers/hs/cmcglone/Student%20Documents/Chapter%204%20(Application%20of%20Derivatives)/Section%204.3%20-%20Olsen%20Curve%20Sketching%20Answers.pdf"
The above link also gives the answer. I am not sure on how to get there.

So, my question is: How does one find the maximum value of a function, on a closed interval, given only the graph of the first derivative?

Homework Equations

Because the question is solely graphical, no equations are needed. However, knowledge of the following theorems are:
Extreme Value Theorem
Rolle's Theorem
Mean Value Theorem

and
The First Derivative Test

The Attempt at a Solution

Well, if I had the equation to f(x), then I could simply plug the critical points into f(x) and the highest value would be the answer.
Perhaps it has something to do with the slope around the point? Or perhaps the answer can be found using a theorem I am not aware of?

OH, -5 is an endpoint! Surely, this must be critical information! But still, how does one determine for sure that this is the maximum value? I know it is a possible candidate.

And that's as far as I get.
Thanks for the help, in advance

 

[flyingpig]

Something isn't right...

if f has a max/min, f' = 0, f'(-5) is NOT 0

 

[://Justice]

Something isn't right...

if f has a max/min, f' = 0, f'(-5) is NOT 0

But it is an endpoint! Endpoints are also possible candidates for maximum values on a closed interval.
I do not know how to determine that it is the maximum value, however.

EDIT:
To clarify, -5 is an endpoint on the closed interval [-5,6]. The question asks for the maximum value on the interval [-5,6].

 

[flyingpig]

Oh okay, my fault.

What is the sign of f'(x) at x = -5?

 

[://Justice]

Oh okay, my fault.

What is the sign of f'(x) at x = -5?

It is negative

 

[flyingpig]

It is negative

And what does that tell you?

 

[://Justice]

And what does that tell you?

That the slope is negative at that point on f(x)

 

[flyingpig]

That the slope is negative at that point on f(x)

Yes, what does that look like on f? When we are going DOWN?

 

[://Justice]

Yes, what does that look like on f? When we are going DOWN?

Umm...
The slope is negative, but is increasing, correct?
Not sure if that was what you were looking for

 

[flyingpig]

Umm...
The slope is negative, but is increasing, correct?
Not sure if that was what you were looking for

Imagine the slope is a ramp. Does that help?

 

[://Justice]

Imagine the slope is a ramp. Does that help?

Ummm... attempting to visualize it, but having some trouble. I do believe that the y value for x=-5 has to be greater than at x=-3

EDIT:
Soo.. the slope is negative at -5, and still negative until 1. It is only positive on the interval (1,3). Logically, it seems that the y value could not go above the y value at -5, because it has decreased to such a degree, and the slope is positive only for a short moment. But I think that is far from mathematical proof.

 

[flyingpig]

Ummm... attempting to visualize it, but having some trouble. I do believe that the y value for x=-5 has to be greater than at x=-3

EDIT:
Soo.. the slope is negative at -5, and still negative until 1. It is only positive on the interval (1,3). Logically, it seems that the y value could not go above the y value at -5, because it has decreased to such a degree, and the slope is positive only for a short moment. But I think that is far from mathematical proof.

Let's try it this way. To get down from a hill, I must climb ______

 

[://Justice]

Let's try it this way. To get down from a hill, I must climb ______

Down, of course
And I can see that x=-5 is a "high" point, but I don't see how to prove that it is the highest

 

[flyingpig]

I hate it when I make a mistake too. But since I messed up, you would probably be able to get the answer.

To get down from a hill, I must have climbed the ____ of the hill first.

 

[://Justice]

I hate it when I make a mistake too. But since I messed up, you would probably be able to get the answer.

To get down from a hill, I must have climbed the ____ of the hill first.

The top of the hill, of course :)
And I understand the there is surely a "top" to the hill, however, how can I prove that it is on the interval [-5,6], and not somewhere farther to the left? How can I tell that x=-5 is not near the bottom of the hill?

 

[flyingpig]

The top of the hill, of course :)
And I understand the there is surely a "top" to the hill, however, how can I prove that it is on the interval [-5,6], and not somewhere farther to the left? How can I tell that x=-5 is not near the bottom of the hill?

You just told me the slope is negative...

 

[://Justice]

Which means that the line is going down. So if you go to the left, you're going up. So the father left, the higher up you are (assuming that there is not another change in slope).
Umm... I'm afraid I'm a bit lost by attempting to visualize the answer. I think I would understand this more easily in terms of math. Sorry I am having trouble with something that must seem obvious

(I am afraid I only have a short time left on the computer)

EDIT:
I have to leave now, but I'll check back later.
Thanks for the help and maybe the solution will come to me!

 

[Ray Vickson]

Homework Statement

The specific problem can be found here: http://www.cbsd.org/sites/teachers/hs/cmcglone/Student%20Documents/Chapter%204%20(Application%20of%20Derivatives)/Section%204.3%20-%20Olsen%20Curve%20Sketching%20Answers.pdf"
The above link also gives the answer. I am not sure on how to get there.

So, my question is: How does one find the maximum value of a function, on a closed interval, given only the graph of the first derivative?

Homework Equations

Because the question is solely graphical, no equations are needed. However, knowledge of the following theorems are:
Extreme Value Theorem
Rolle's Theorem
Mean Value Theorem

and
The First Derivative Test

The Attempt at a Solution

Well, if I had the equation to f(x), then I could simply plug the critical points into f(x) and the highest value would be the answer.
Perhaps it has something to do with the slope around the point? Or perhaps the answer can be found using a theorem I am not aware of?

OH, -5 is an endpoint! Surely, this must be critical information! But still, how does one determine for sure that this is the maximum value? I know it is a possible candidate.

And that's as far as I get.
Thanks for the help, in advance

It is possible to cook up examples where knowing f' alone is not enough. This may be the case when, for example, we have end-point minima, with f'(a) > 0 at the left end and f'(b) < 0 at the right end. Both x=a and x=b are *local* minima, but we need to actually compute f(a) and f(b) (or f(b) - f(a)) in order to tell which is the true minimum. In other words, we may need to accurately estimate the integral of f'(x) for x from a to b in order to fully answer the question. Of course, not all examples are like that.

RGV