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Math Help - h/w due for 2morrow

  1. #1
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    h/w due for 2morrow

    prove for k is bigger than or equal to 0.

    on attachment!
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  2. #2
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    Quote Originally Posted by ardam View Post
    prove for k is bigger than or equal to 0.

    on attachment!
    Consider f(x) = x^{1/3}.

    Then f(x + h) \leq f(x) + h f'(x) for h > 0 and x > 0.
    Last edited by mr fantastic; January 13th 2009 at 11:20 AM. Reason: Added x > 0
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Consider f(x) = x^{1/3}.

    Then f(x + h) \leq  f(x) + h f'(x) for h > 0.
    Smart alec! LOL. I knew I was missing something obvious. I tried so many algebraic manipulations it was driving me nuts! Hahaha
    Last edited by mr fantastic; January 13th 2009 at 11:20 AM. Reason: I had a posterior reason
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    Quote Originally Posted by mr fantastic View Post
    Consider f(x) = x^{1/3}.

    Then f(x + h) \leq  f(x) + h f'(x) for h > 0.
    thanks for help but not sure why you differentiate.
    Last edited by mr fantastic; January 13th 2009 at 11:20 AM.
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    Quote Originally Posted by ardam View Post
    thanks for help but not sure why you differentiate.
    It's a standard inequality.

    Use f(x) = x^{\frac{1}{3}} and h = 1.

    When you sub everything into the inequality you should get

    (x + 1)^{\frac{1}{3}} \leq x^{\frac{1}{3}} + \frac{1}{3}x^{-\frac{2}{3}}.
    Last edited by mr fantastic; January 13th 2009 at 11:21 AM.
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    Quote Originally Posted by Prove It View Post
    It's a standard inequality.

    Use f(x) = x^{\frac{1}{3}} and h = 1.

    When you sub everything into the inequality you should get

    (x + 1)^{\frac{1}{3}} \leq x^{\frac{1}{3}} + \frac{1}{3}x^{-\frac{2}{3}}.
    oh right yer i previously got that far but not sure wot technique to use to prove.
    Last edited by mr fantastic; January 13th 2009 at 11:21 AM.
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  7. #7
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    Quote Originally Posted by ardam View Post
    oh right yer i previously got that far but not sure wot technique to use to prove.
    Substitute x = k and make the obvious re-arrangement.
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    Quote Originally Posted by mr fantastic View Post
    Substitute x = k and make the obvious re-arrangement.
    so then you prove its bigger than of equal to 0
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    Quote Originally Posted by ardam View Post
    so then you prove its bigger than of equal to 0
    This inequality is known to be true.

    So make the appropriate substitutions and rewrite the fractional powers as surds, then move the \sqrt[3]{k} to the other side.

    What do you get?
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    Quote Originally Posted by Prove It View Post
    This inequality is known to be true.

    So make the appropriate substitutions and rewrite the fractional powers as surds, then move the \sqrt[3]{k} to the other side.

    What do you get?
    you get that all the terms are less than 0
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  11. #11
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    Quote Originally Posted by ardam View Post
    you get that all the terms are less than 0
    Surely you can re-arrange the inequality I gave you (and there's a small edit I made) into the form of the inequality you're required to prove. This is the only bit of the solution that has been left for you to do.
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