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Thread: 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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    $\displaystyle \frac{5x^2-7x+2}{4x^2-2}$

    We can try to make the denominator 0.

    $\displaystyle 4x^2-2=0$

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

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

    Now, plug those values in the function and check which one makes it $\displaystyle +\infty$ or $\displaystyle - \infty$
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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
    $\displaystyle f(x) = \frac{5x^2 - 7x + 2}{4x^2 - 2}$.

    Divide numerator and denominator by $\displaystyle x^2$:

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

    Therefore:

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

    We can try to make the denominator 0.

    $\displaystyle 4x^2-2=0$

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

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

    Now, plug those values in the function and check which one makes it $\displaystyle +\infty$ or $\displaystyle - \infty$
    You are making $\displaystyle \mathrm{f}(x)\to\pm\infty$. The question is about $\displaystyle x\to\pm\infty$.
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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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