1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Min/max functions in single-variable calc

  1. Dec 2, 2007 #1
    [SOLVED] Min/max functions in single-variable calc

    1. The problem statement, all variables and given/known data
    7. If x is positive and m greater than one, prove that x^m - 1 - m(x-1) is not negative using max-min.




    2. Relevant equations



    3. The attempt at a solution

    I can't figure out a secondary equation to get the original in terms of x or m so I can do the first and second derivatives. There must be something stupid I'm missing... I don't know if you help in the situations where the first step is needed o_O
     
  2. jcsd
  3. Dec 2, 2007 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    With [itex]f(x)=x^m-1-m(x-1)[/itex], where does f(x) reach extremum?
     
  4. Dec 2, 2007 #3
    At m = 1, f(x) = 0 regardless. When m is any greater than that f(x) is greater than 0 (since x^m increases exponentially whereas m(x-1) increases linearly). But I don't know how to show this without simply giving data; it has to be mathematical.
     
  5. Dec 2, 2007 #4
    I guess my problem is more in conveying the information to my teacher than actually solving the problem....
     
  6. Dec 2, 2007 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You didn't answer my question. What level math are you in? Do you know about using derivatives to find extrema?
     
    Last edited: Dec 2, 2007
  7. Dec 2, 2007 #6
    Calc BC (high school level), and yes. I figured out what I was doing wrong; m is a constant and can be treated as thus. Doing the limit that you suggested before editing that post was also a bit misleading.

    Since m is constant, you can take the derivative of the original equation and get f'(x) = mx^(m-1) - m. Set it equal to 0, you end up with mx^(m-1)=m, them x^(m-1) = 1.

    Solve for x, you get x^m - x = 0, or x^m = x.

    Since m > 1 and thus cannot be 1, and x > 0 and cannot be 0, x must = 1.

    Go back to the original equation f(x) = 1^m - 1 - m(x - 1)

    Simplify to 1^m - 1. For any value of m, 1^m will always equal 1, thus the equation always = 0 which isn't negative.

    Which begs the question, ARE there any extrema in this equation? It's always 0....

    Edit: Oh, and thanks as well. I did have my "Oh, duh!" moment while looking at this post, so....
     
    Last edited: Dec 2, 2007
  8. Dec 2, 2007 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    The derivative is always zero because you forced it to be so by choosing x=1. The original function, [itex]f(x)=x^m-1-m(x-1)[/itex] is not identically zero. For example, [itex]f(0)=m-1[/itex]. The derivative is zero at x=1 for all m. This does not mean the derivative is identically zero for all x.

    Do you know about the second derivative test to determine whether an extremum is a minimum or a maximum?
     
  9. Dec 2, 2007 #8
    Yes, I do. if f '' (x) > 0, x is a min, opposite it's a max, and if f '' (x) = 0 it's a Point of Inflection (usually anyway, for some reason (something that came up earlier in the assignment) in a cosine function my min had an f '' (x) of 0).

    I'm not catching something. The second derivative would be (m-1)(m)x^(m-2). If m > 1 and x > 0, then the second derivative must be (+) (+) (+)^(+/-), overall becoming positive, simply proving that this is a minimum. But I don't see how one can go from there to prove that that minimum is greater than 0 in f(x) without "forcing" it to be 0.
     
  10. Dec 2, 2007 #9
    Oh, is that really forcing the equation to be 0? We're supposed to work with critical numbers, and the only one (that I found anyway) was x = 1, given all of the criteria.

    I'm really sorry if I'm like making you bash your head into your desk for not noticing something painfully obvious.
     
  11. Dec 2, 2007 #10

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    First, a minor correction on post #6.
    Correct.
    Correct.
    Why did you do this? Doing this introduces zero as a solution, which it isn't since [itex]m>1[/itex]. One is obviously a solution to [itex]x^{m-1}=1[/itex]. The only other possible real root to the expression [itex]x^{m-1}=1[/itex] is [itex]x=-1[/itex], and you can ignore this possibility since [itex]x>0[/itex].


    To summarize, this is what you have deduced
    • The function has one and only one extremum for positive x, and this is at x=1 (post #6)
    • This one extremum is a minimum (post #8).

    You need just one more piece of information to finish the problem. Hint: What is the minimum value of the function?
     
  12. Dec 2, 2007 #11
    ze...ro? Just prove that the minimum of this function for positive x, x = 1, is 0 which proves that everything else is greater than zero, ergo positive? Didn't I do that in post 6?

    "Simplify to 1^m - 1. For any value of m, 1^m will always equal 1, thus the equation always = 0 which isn't negative."

    I might have worded it incorrectly, I should have stated that the minimum is 0, but is that the last piece of needed info?

    And on the introducing 0 as an option, x can't equal 0 because x is positive (>0). I...don't know why I did it that way either, to tell you the truth. It was just easier to look at I think.
     
  13. Dec 2, 2007 #12

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You have shown the minimum value the function can obtain for any positive x is zero. Now look at the problem statement itself: "If x is positive and m greater than one, prove that x^m - 1 - m(x-1) is not negative". There is only one extremum, and it is a minimum. This minimum is zero. How can the function be negative?

    Suppose the function does have a negative value for some positive x, call it x0. This means the function must have a local maximum somewhere between x=1 and x=x0. But it doesn't because the function has only one extremum.
     
  14. Dec 2, 2007 #13
    Alright. Danke Schoen D H.
     
  15. Dec 2, 2007 #14

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You're welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?