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Determining if a function has a cusp, vertical tangent, corner or none at a point

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    At x=3 does the function y=x^3 + (x-3)^(1/3) have a vertical tangent, cusp, corner or none?

    3. The attempt at a solution

    I took the derivative

    y'=3x^2 + 1 / 3x^(2/3) then i replaced 3 in which gave 27 + 1 / 0

    I don't understand how to come to the correct conclusion following this step.

    Vertical tangent comes to mind since 1 / 0 is a vertical line, but I don't know how to prove it using limits.
  2. jcsd
  3. Jan 24, 2010 #2


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    You mean y'=3x^2 + 1 / 3(x-3)^(2/3), right? If you approach x=3 from below then y' goes to +infinity. I you approach it from above it also goes to +infinity. y is also continuous at x=3. Looks like a vertical tangent to me.
  4. Jan 24, 2010 #3


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    Is that really what you got?? It's either a typo or a mis-calculation.

    If you have a positive infinite limit from both the left right that suggests a vertical line alright. You can tell whether it is vertical tangent line or cusp by looking at concavity on each side of x = 3.
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