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Math Help - show that...what method?

  1. #1
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    show that...what method?

    By considering \frac{1}{3}x - x^{\frac{1}{3}} or otherwise, show that:

    \sqrt[3]{x} < \frac{1}{3}x + \frac{2}{3} for all x with 0 < x < 1.


    This is another past paper anaylsis question...just wondering if someone could please tell me what method should be used to do this?! I have no idea! Possibly Taylor series?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by hunkydory19 View Post
    By considering \frac{1}{3}x - x^{\frac{1}{3}} or otherwise, show that:

    \sqrt[3]{x} < \frac{1}{3}x + \frac{2}{3} for all x with 0 < x < 1.


    This is another past paper anaylsis question...just wondering if someone could please tell me what method should be used to do this?! I have no idea! Possibly Taylor series?

    Thanks in advance!
    What I would do is multiply both sides of the inequality by 3 and then cube both sides, which maintains the inequality. That gives you 27x < (x + 2)^3, or 27x < x^3 + 6x^2 + 12x + 8. Finally, subtracting 27x from both sides gives you 0 < x^3 + 6x^2 -15x + 8. You can find the roots of the equation x^3 + 6x^2 - 15x + 8 = 0 and then analyze the interval x \in (0, 1). Note that 1 is a root of this equation, which should make it easier to factor.
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