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Finding realtive extrema

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Let h:(0,1)→ℝ be defined by h(x)=cos([itex]\pi\\[/itex]/x). Determine the set of all x such that x has a relative extreme value at x.



    2. Relevant equations



    3. The attempt at a solution
    It is common knowledge that the min and max are -1 and 1. Using intuition, it is clear by guess and check to arrive at x=1/(2n+1) gives a minimum and x=1/2n gives a maximum.

    Using the first derivative test, I arrive at 1=0 which is not true. I am unsure how to interpret this answer.

    Trying to solve [itex]d^{2}[/itex](cos([itex]\pi\\[/itex]/x))/d[itex]x^{2}[/itex]=0 is proving to be a real challenge.

    I am only concerned about this since my professor wants me to show my work, so I do not think saying I guessed and checked will be acceptable.
     
  2. jcsd
  3. Mar 11, 2012 #2

    Mark44

    Staff: Mentor

    Instead of guessing at the answer, find the values of x for which h'(x) = 0.

    Show us what you did to arrive at 1 = 0.
     
  4. Mar 11, 2012 #3
    Using the chain rule, h'(x)=[itex]\pi\\[/itex]sin([itex]\pi\\[/itex]/x)*1/x[itex]^{2}[/itex]

    Set it equal to 0

    [itex]\pi\\[/itex]sin([itex]\pi\\[/itex]/x)*1/x[itex]^{2}[/itex]=0

    Multiply both sides by x[itex]^{2}[/itex]/[itex]\pi\\[/itex] yields sin([itex]\pi\\[/itex]/x)=0

    Take the arcsin of both sides yields

    [itex]\pi\\[/itex]/x=0

    Divide both sides by pi,

    1/x=0

    Multiply both sides by x

    1=0

    Or am I forgetting that there are more possible values that arcsin can be 0?
     
  5. Mar 11, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You should not be looking at the arcsin; you should be looking for solutions of the equation sin(w) = 0. Can you see why the arcsin misses all but one solution?

    RGV
     
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