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Inequality Maximization

  1. Jan 14, 2014 #1
    1. |(x2+2x+1)/(x2+3)|≤ M. Find the value of M when |x|≤ 3.



    2. |u+v|=|u|+|v|



    3. I understand that you start off by distributing the absolute value symbols into the individual terms as above. Then you maximize the numerator, using 3 as the value for x. However, my professor then minimized the denominator by using 0 as the value for x. This makes M = 240/3. What I do not understand is how you can use 2 different values of x to determine one value for M. Is there a value of x then that gives you M as the outcome in the L.H.S.?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 14, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Is x2 supposed to be x2? If so, either use the palette at the top of the input screen (as I just now did), or else write x^2.

    Your statement |u+v| = |u| + |v| is false! For example, 0 = |1 + (-1)| ≠ |1| + |-1| = 2.
     
  4. Jan 14, 2014 #3

    Mark44

    Staff: Mentor

    As Ray already mentioned, your equation |u + v| = |u| + |v| is false.

    The absolute values in your inequality can be removed, because x2 + 2x + 1 ≥ 0 for all real x, and x2 + 3 > 0 for all real x.
     
  5. Jan 14, 2014 #4
    I knew something was wrong with this question. Thank you for your replies, I will ask my professor for further clarification.
     
  6. Jan 14, 2014 #5

    Mark44

    Staff: Mentor

    Why? I don't think there's anything wrong with the question. Our complaint was with what you wrote as a relevant equation.

    After doing some simplification, you wind up with a quadratic inequality.
     
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