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find the domain and range of the function f(x)=x^2-9/4x^2+x

  • math/calculus -

    You have

    (x-3)(x+3) / x(4x+1)

    The domain is all real numbers except where the denominator is zero. So, that would be all reals except 0 and -1/4

    There is a horizontal asymptote at y = 1/4.
    There are vertical asymptotes at x = -1/4 and x=0.

    For x between -1/4 and 0, y > 0
    As x gets big, y lies below the asymptote.

    So, the range outside (-1/4,0) is all reals less than 1/4.

    y' = (x^2+72x+9)/(4x^2+1)^2
    y' = 0 when x = -.125
    y(-.125) = 143.75

    So, the range is all reals
    < 1/4 or >= 143.75

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