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Math Help - How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

  1. #16
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    Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

    Hmm...alternatively you can let x = \frac{\cot \theta}{\sqrt{3}}, you should still obtain the same thing.
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  2. #17
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    Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

    It's okay, since mostly, we use diffentiation while tri sub. is not a always useful method instead. Don't feel bad for not thinking for tri method, it's okay
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  3. #18
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    Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

    Daigo has just taken Algebra. Soon, i think the suggests should somewhat easier to understand; not calculus or trigonometry, but basic Algebra.

    You should estimate the points on the graph that are max. and min.
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  4. #19
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    Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

    Aha! I thought there would be an non-calculus, non-trig, algebraic solution to this problem.


    Let k be the maximal value of \frac{2x}{3x^2 + 1} so that \frac{2x}{3x^2 + 1} = k. Cross-multiplying, we obtain


    3kx^2 - 2x + k = 0 \Rightarrow x = \frac{2 \pm \sqrt{4 - 4(3k)(k)}}{6k} = \frac{2 \pm \sqrt{4 - 12k^2}}{6k}. We want k to be as large as possible subject to the constraint that x is real. This implies that

    4 - 12k^2 \ge 0 \Rightarrow k^2 \le \frac{1}{3}. The largest possible value of k is \sqrt{\frac{1}{3}}.

    Furthermore, since y = \frac{2x}{3x^2 + 1} is an odd function, the smallest possible value of k is -\sqrt{\frac{1}{3}}.
    Last edited by richard1234; June 15th 2012 at 09:03 PM.
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