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Posted by on Wednesday, March 18, 2009 at 3:43pm.

f(x)= x^3/x^2-16 defined on [-19, 16]

How would you find the vertical asymptopes? and the inflection point?

I think you take the derivative and set it to zero? but i cant figure out the correct derivative or something??

  • calc - , Wednesday, March 18, 2009 at 3:51pm

    assume you mean (x^2-16) on the bottom

    vertical asymptotes are where the denominator goes to zero
    that is when x = -4 and x = +4

    derivative of fraction =
    (bottom * derivative of top - top * derivative of bottom)
    all over bottom squared

    [(x^2-16)*3 x^2 - x^3(2x) ]/bottom^2
    look for zeros of numerator
    3 x^4 - 48 x^2 - 2 x^4
    =x^4 - 48 x^2
    = x^2 (x^2-48)
    = x^2 (x-sqrt 48)(x+sqrt 48)
    so zero at -sqrt 48, 0 and + sqrt 48
    check my arithmetic, I did that fast.

  • calc - , Wednesday, March 18, 2009 at 3:52pm

    by the way sqrt 48 = 4 sqrt 3

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