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Mean value thereom application

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose that x,y are in (0,0.5). Show that:
    $$\left|\frac{1}{x^3}-\frac{1}{y^3}\right| ≥ 48|x-y|$$

    2. Relevant equations


    3. The attempt at a solution
    I think that we need a function ##f(t) = t^{-3}## & ##f'(t) = -3t^{-4}##. Now we use the MVT here to show that there is some c that makes the expression in the problem correct right? My question is how do I incorporate the absolute value notations in the statement here?
    $$-3c^{-4} = \frac{f(y) - f(x)}{y-x}$$
    I think we have to assume that either x or y is greater than the other(because if they were equal then the above owuld be proven). Do I just input c for 0.5? I'm unsure.
     
  2. jcsd
  3. Dec 10, 2016 #2

    PeroK

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    It's not clear why the MVT would help here.
     
  4. Dec 10, 2016 #3
    I forgot to add that they are differentiable on the interval (x,y)
     
  5. Dec 10, 2016 #4

    PeroK

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    I still can't see why you think the MVT helps you.
     
  6. Dec 10, 2016 #5

    LCKurtz

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    Remember that all your variables are in ##(0,\frac 1 2)##. If you take the absolute value of both sides of what you have written you get$$
    \frac 3 {c^4} =\frac{|f(y) - f(x)|}{|y-x |}$$which can be rewritten$$
    \frac 3 {c^4} |y-x| =\left | \frac 1 {x^3} - \frac 1 {y^3}\right |$$Think about how large or small the left side can be.
     
  7. Dec 12, 2016 #6
    Since c is an element of x,y then it can't be larger than 0.5 or smaller than 0? And as such this proves that since 48|y-x| will be greater since y is greater than c? So the side with c will always be smaller.
     
    Last edited: Dec 12, 2016
  8. Dec 12, 2016 #7

    LCKurtz

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    That doesn't make any sense. ##c## is a number in ##(0,.5)##, not an "element" of x,y, whatever that means.

    Stating the result doesn't prove anything. You need to show some inequalities.
     
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