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Calculus 1: Finding the maxima and minima of the function 1-x^(2/3)

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data

    I am given the problem "A function is defined by f(x) = 1-x^(2/3).

    a. find the minima and maxima of the function. State whether they are relative or absolute.

    b. graph the function."


    2. Relevant equations

    I found the derivative and set it equal to zero. y' = -(2/3)x^(-1/3), -(2/3)x^(-1/3) = 0


    3. The attempt at a solution

    Now I really have little idea of where to go from here because I am left with a few questions. If i set the derivative equal to zero and set x to be zero would the statement "-(2/3)0^(-1/3) = 0" be true or would I essentially be dividing by zero because x^(-1/3) = 1/ [itex]\sqrt[3]{x}[/itex]? Now if this is a spot where the velocity is zero would you consider it a max, a min, or would it be irrelevant to my answer because you can't quite classify it either way. I think it is simply an interesting value on the graph, but is neither a maximum or minimum relative or absolute, and this graph has no maxes or mins.

    Thanks for your help. I really appreciate it.
     
  2. jcsd
  3. Apr 1, 2012 #2

    Dick

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    Sketch a graph. You are correct that the derivative at x=0 doesn't exist. It still might be a max or a min. It's an "interesting point" as you put it.
     
  4. Apr 2, 2012 #3
    Thanks again.
     
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