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Calc I

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y = (x^2 -4)/(x^2 - x -12)

dy/dx of numerator I got 2x
dy/dx of denomenator I got 2x - 1
dy/dx of y I got (x^2 - 16x -4)/((x-4)^2(x+3)^2)

setting numerator to zero and solving got 8 +/- sqrt(65) with the critical value being the +
setting denomenator to zero and solving got 4 and -3 which are both out of range of the original function...

so now what how do I find the range kind of confused... found one critical value how do I find the others kind of lost here thanks

  • Calc I -

    What are you trying to do with this question?
    are you finding dy/dx ?
    Using the quotient rule your dy/dx should have been

    -(x^2 + 16x + 4)/(x^2-x-12)^2

    setting that equal to zero would give you the x's of the max/mins.
    x^2 + 16x + 4 = 0
    x = (-16 ±√(240)/2
    = -8 ± √60

    since the denominator factors to (x-4)(x+3) there are two vertical asymtotes, x = 4 and x = -3
    so the range would be -infinitiy to + infinity
    since the curve will rise and fall along these vertical lines

    Why did you simply take the derivative of the numerator and the denominator?
    Is this some king of "creative math" or are you perhaps doing L'Hopital's Rule and attempting to find the Limit ?

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