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

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    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" [Broken]
    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?

    2. Relevant 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

    The First Derivative Test

    3. 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
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 22, 2011 #2
    Something isn't right...

    if f has a max/min, f' = 0, f'(-5) is NOT 0
  4. Sep 22, 2011 #3
    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.

    To clarify, -5 is an endpoint on the closed interval [-5,6]. The question asks for the maximum value on the interval [-5,6].
  5. Sep 22, 2011 #4
    Oh okay, my fault.

    What is the sign of f'(x) at x = -5?
  6. Sep 22, 2011 #5
    It is negative
  7. Sep 22, 2011 #6
    And what does that tell you?
  8. Sep 22, 2011 #7
    That the slope is negative at that point on f(x)
  9. Sep 22, 2011 #8
    Yes, what does that look like on f? When we are going DOWN?
  10. Sep 22, 2011 #9
    The slope is negative, but is increasing, correct?
    Not sure if that was what you were looking for
  11. Sep 22, 2011 #10
    Imagine the slope is a ramp. Does that help?
  12. Sep 22, 2011 #11
    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

    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.
  13. Sep 22, 2011 #12
    Let's try it this way. To get down from a hill, I must climb ______
  14. Sep 22, 2011 #13
    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
  15. Sep 22, 2011 #14
    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.
  16. Sep 22, 2011 #15
    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?
  17. Sep 22, 2011 #16
    You just told me the slope is negative...
  18. Sep 22, 2011 #17
    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)

    I have to leave now, but I'll check back later.
    Thanks for the help and maybe the solution will come to me!
    Last edited: Sep 22, 2011
  19. Sep 22, 2011 #18

    Ray Vickson

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    Science Advisor
    Homework Helper

    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.

    Last edited by a moderator: May 5, 2017
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