Results 1 to 4 of 4

Thread: Find the maximum/minimum

  1. #1
    Senior Member
    Joined
    Nov 2007
    Posts
    329

    Find the maximum/minimum

    Find the local maximum and local minimum of $\displaystyle f(x)=(x-1)(x-2)(x-3)$ using AM-GM inequality.
    Last edited by james_bond; Apr 30th 2008 at 08:14 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,410
    Thanks
    1
    f(x) tends to negative and positive infinity doesn't it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    Let’s say we want to find the maximum and minimum of f(x) in the interval $\displaystyle 1\leq x\leq3$. That would make much more sense, wouldn’t it?

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

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

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

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

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

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

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

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

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

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

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

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

    Hence the maximum and minimum of f(x) are $\displaystyle \frac{2\sqrt{3}}{9}$ and $\displaystyle -\,\frac{2\sqrt{3}}{9}$, attained when $\displaystyle x=2-\frac{1}{\sqrt{3}}$ and $\displaystyle x=2+\frac{1}{\sqrt{3}}$ respectively.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the minimum & maximum area of a box!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 13th 2011, 12:51 PM
  2. find absolute minimum and maximum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 2nd 2009, 03:26 PM
  3. Replies: 2
    Last Post: Sep 10th 2009, 04:27 PM
  4. How to find Maximum and Minimum Points?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jul 26th 2009, 02:49 AM
  5. Find maximum and minimum value of
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 26th 2009, 08:53 PM

Search Tags


/mathhelpforum @mathhelpforum