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Thread: inequality

  1. #1
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    inequality

    Find the set of real values of x for which :

    $\displaystyle \frac{2}{x} \geq 3-x$

    I got x=-3 , x=1

    what should I do next ?
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  2. #2
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    Quote Originally Posted by Musab View Post
    Find the set of real values of x for which :

    $\displaystyle \frac{2}{x} \geq 3-x$

    I got x=-3 , x=1

    what should I do next ?
    Solve the following cases:

    Case 1: $\displaystyle x > 0$: $\displaystyle 2 \geq 3x - x^2 \Rightarrow x^2 - 3x + 2 \geq 0$.

    Case 2: $\displaystyle x < 0$: $\displaystyle 2 \leq 3x - x^2 \Rightarrow x^2 - 3x + 2 \leq 0$.

    Edit: Note that $\displaystyle x \neq 0$ (why?)
    Last edited by mr fantastic; Sep 6th 2008 at 04:43 PM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Solve the following cases:

    Case 1: $\displaystyle x > 0$: $\displaystyle 2 \geq 3x - x^2 \Rightarrow x^2 - 3x + 2 \geq 0$.

    Case 2: $\displaystyle x < 0$: $\displaystyle 2 \leq 3x - x^2 \Rightarrow x^2 - 3x + 2 \leq 0$.
    Case 1: $\displaystyle x^2 - 3x + 2 \geq 0 \Rightarrow$ $\displaystyle x \leq 1$ or $\displaystyle x \geq 2$ subject to the restriction $\displaystyle x > 0$. Therefore $\displaystyle x \geq 2$ or $\displaystyle 0 < x \leq 1$.

    Case 2: Left for you.
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