1. Limited time only! Sign up for a free 30min personal 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!

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


    User Avatar
    Science Advisor
    Homework Helper

    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(\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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook