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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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    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

    Staff: Mentor

    First, a minor correction on post #6.
    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

    Staff: Mentor

    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

    Staff: Mentor

    You're welcome.
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