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Homework Help: Mean Value Theorem exercise (Analysis)

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [tex]a>b>0[/tex] and let [tex]n \in \mathbb{N}[/tex] satisfy [tex]n \geq 2[/tex]. Prove that [tex]a^{1/n} - b^{1/n} < (a-b)^{1/n}[/tex].
    [Hint: Show that [tex]f(x):= x^{1/n}-(x-1)^{1/n}[/tex] is decreasing for [tex]x\geq 1[/tex], and evaluate [tex]f[/tex] at 1 and a/b.]

    2. Relevant equations

    I assume, since this exercise is at the end of the Mean Value Theorem section, I am to use the Mean Value Theorem.

    3. The attempt at a solution

    I can show what the hint suggests. I guess I'm not sure how those ideas help exactly.
     
  2. jcsd
  3. Nov 25, 2007 #2

    morphism

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    Did you evaluate f at 1 and a/b?

    After you do that, keep in mind that a>b>0. You don't need to explicitly apply the mean value theorem (but you probably implicitly applied it when you proved that f(x) was decreasing for x>=1).
     
  4. Nov 25, 2007 #3
    Thanks for your quick reply. Let's see what we have here....

    [tex]f(1)=1[/tex]

    and

    [tex]f(\frac{a}{b}) = \frac{a^{1/n}-(a-b)^{1/n}}{b^{1/n}}[/tex].
     
    Last edited: Nov 26, 2007
  5. Nov 26, 2007 #4
    ohh, i think i see now.

    [tex] 1 < a/b[/tex]

    and, since f is decreasing,

    [tex]f(\frac{a}{b}) = \frac{a^{1/n}-(a-b)^{1/n}}{b^{1/n}} < 1 = f(1)[/tex]

    and the rest is just algebra to show

    [tex]a^{1/n} - b^{1/n} < (a-b)^{1/n}[/tex].

    look good?
     
    Last edited: Nov 26, 2007
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