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  1. #1
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    function problem

    Hello! I have this problem.

     f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}
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    Quote Originally Posted by Anemori View Post
    Hello! I have this problem.

     f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}
    Solve \frac {x^2}{x^2+1} = \frac {1}{2} .... (1)

    Then draw a graph of y = \frac {x^2}{x^2+1} = - \frac{1}{x^2 + 1} + 1 and use it and the solutions to equation (1) above to solve the given inequation.
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  3. #3
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    Quote Originally Posted by Anemori View Post
    Hello! I have this problem.

     f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}
    Here is a slightly different approach:

    Since x^2 + 1 \geq 1 the inequality becomes:

    x^2\geq\frac12 x^2 + \frac12

    \frac12 x^2- \frac12\geq 0~\implies~x^2-1\geq 0~\implies~(x+1)(x-1)\geq 0

    A product of 2 factors is positive if both factors have the same sign:

    x+1\geq 0\wedge x-1\geq 0~\vee ~x+1\leq 0\wedge x-1\leq 0

    x\geq -1\wedge x\geq 1~\vee ~x\leq -1\wedge x\leq 1

     x\geq 1~\vee ~x\leq -1
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  4. #4
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    Quote Originally Posted by Anemori View Post
    Hello! I have this problem.

     f(x) = \frac {x^2}{x^2+1} >= \frac {1}{2}
    Rewrite it as \frac{x^2}{x^2+1}- \frac{1}{2}= \frac{2x^2}{2(x^2+1}- \frac{x^2+ 1}{2(x^2+ 1}= \frac{x^2- 1}{x^2+ 1}\ge 0

    A fraction is positive if and only if the numerator and denominator have the same sign. This, denominator, x^2+ 1, is always positive.
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