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Math Help - Little help please

  1. #1
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    Little help please

    f(x)=(5x^2-7x+2)/(4x^2-2)

    f(x)--> ? as x--> -infinity

    f(x)--> ? as x--> infinty

    can someone help me out here, i seem to be coming up with the wrong answers..thanks
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  2. #2
    Super Member wingless's Avatar
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    \frac{5x^2-7x+2}{4x^2-2}

    We can try to make the denominator 0.

    4x^2-2=0

    x^2 = \frac{1}{2}

    x = \{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \}

    Now, plug those values in the function and check which one makes it +\infty or - \infty
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  3. #3
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    Quote Originally Posted by mathlete View Post
    f(x)=(5x^2-7x+2)/(4x^2-2)

    f(x)--> ? as x--> -infinity

    f(x)--> ? as x--> infinty

    can someone help me out here, i seem to be coming up with the wrong answers..thanks
    f(x) = \frac{5x^2 - 7x + 2}{4x^2 - 2}.

    Divide numerator and denominator by x^2:

    f(x) = \frac{5 - \frac{7}{x} + \frac{2}{x^2}}{4 - \frac{2}{x^2}}.

    Therefore:

    As x \rightarrow \pm \infty, f(x) \rightarrow \frac{5 - 0 + 0}{4 - 0} = \frac{5}{4}.
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by wingless View Post
    \frac{5x^2-7x+2}{4x^2-2}

    We can try to make the denominator 0.

    4x^2-2=0

    x^2 = \frac{1}{2}

    x = \{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \}

    Now, plug those values in the function and check which one makes it +\infty or - \infty
    You are making \mathrm{f}(x)\to\pm\infty. The question is about x\to\pm\infty.
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  5. #5
    Super Member wingless's Avatar
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    Haha sorry I just misread it
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