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Prove that m<f(x)<M

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    prove that m<f(x)<M
    so [itex]\forallx[/itex][itex]\in[/itex]R: f(x)=[itex]\frac{x}{x^2 +x+1}[/itex]
    the question is prove that m[itex]\leq[/itex]f(x)[itex]\leq[/itex]M
    M and m are real numbers
    2. Relevant equations



    3. The attempt at a solution
    all i did so far was making f(x) : (x+1)^2 +(3/4) well i noticed that m=-1 and M=4/3 from the graph but i can't really prove it well
    f(x)= 1-[itex]\frac{x^2+1}{x^2 +x +1}[/itex] = 1-[itex]\frac{x^2+1}{(x+1/2)^2+3/4}[/itex]
     
  2. jcsd
  3. Dec 13, 2012 #2
    If you know the values of m = -1 and M = 4/3, you can try a direct approach:

    If you want to prove ##-1 \leq \frac x{x^2+x+1}##, you can multiply both sides of that inequality by ##(x^2+x+1) = (x+\frac12)^2+\frac34##, which is positive (and so will not change the direction of the ##\leq## relation symbol):
    $$
    -(x^2+x+1) \leq \frac{x(x^2+x+1)}{x^2+x+1}.
    $$
    Can you now prove that this inequality is true?
     
  4. Dec 13, 2012 #3
    well there is a slight problem here the exercise dosen't mention the -1 nor the 4/3
    so they are expecting us to do it "manually".
     
  5. Dec 13, 2012 #4

    pasmith

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    Homework Helper

    One approach to this sort of problem is to ask "What conditions must [itex]y[/itex] satisfy for [itex]y = f(x)[/itex] to have real solutions for [itex]x[/itex]?"

    This is a particularly good approach in this case, because if [itex]y = x/(x^2 + x + 1)[/itex] then
    [tex]yx^2 + (y-1)x + y = 0[/tex]
    and hopefully you know the condition for that quadratic in [itex]x[/itex] to have real roots. That will give you a condition which [itex]y[/itex] must satisfy, which in turn will give you bounds for [itex]f(x)[/itex].
     
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