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Math Help - Find the maximum/minimum

  1. #1
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    Find the maximum/minimum

    Find the local maximum and local minimum of f(x)=(x-1)(x-2)(x-3) using AM-GM inequality.
    Last edited by james_bond; April 30th 2008 at 08:14 AM.
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  2. #2
    o_O
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    f(x) tends to negative and positive infinity doesn't it?
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  3. #3
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    Quote Originally Posted by o_O View Post
    f(x) tends to negative and positive infinity doesn't it?
    Perhaps the question should read: find the local minimum and maximum, or find the minimum and maximum on an interval.
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  4. #4
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    Let’s say we want to find the maximum and minimum of f(x) in the interval 1\leq x\leq3. That would make much more sense, wouldn’t it?

    Now, rewrite the function as \mathrm{f}(x)=(x-1)(x-2)(x-3)=(x-2)^3-(x-2).

    Case 1: \color{white}.\ . 2\leq x\leq3

    Then x-2\geq0 and by AM–GM, we have

    \color{white}.\quad. \frac{(x-2)^3+\left(\frac{1}{\sqrt{3}}\right)^3+\left(\frac  {1}{\sqrt{3}}\right)^3}{3}\ \geq\ \frac{x-2}{3}

    \Rightarrow\ (x-2)^3-(x-2)\ \geq\ -2\left(\frac{1}{\sqrt{3}}\right)^3=-\,\frac{2\sqrt{3}}{9}

    equality being attained if and only if x-2=\frac{1}{\sqrt{3}}, i.e. if and only if x=2+\frac{1}{\sqrt{3}}

    Case 2: \color{white}.\ . 1\leq x\leq2

    Now x-2\leq0 and so we must use 2-x\geq0 in the AM–GM formula.

    \color{white}.\quad. \frac{(2-x)^3+\left(\frac{1}{\sqrt{3}}\right)^3+\left(\frac  {1}{\sqrt{3}}\right)^3}{3}\ \geq\ \frac{2-x}{3}

    \Rightarrow\ (2-x)^3-(2-x)\ \geq\ -2\left(\frac{1}{\sqrt{3}}\right)^3=-\,\frac{2\sqrt{3}}{9}

    \Rightarrow\ (x-2)^3-(x-2)\ \leq\ \frac{2\sqrt{3}}{9}

    Again equality is attained if and only if 2-x=\frac{1}{\sqrt{3}}, i.e. if and only if x=2-\frac{1}{\sqrt{3}}

    Hence the maximum and minimum of f(x) are \frac{2\sqrt{3}}{9} and -\,\frac{2\sqrt{3}}{9}, attained when x=2-\frac{1}{\sqrt{3}} and x=2+\frac{1}{\sqrt{3}} respectively.
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