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Critical points

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine function [tex] f(x)=\frac{\sqrt[3]{x-4}}{x-1}[/tex] critical points and find max and min value in given interval [tex] [2; 12] [/tex]

    3. The attempt at a solution

    1) I've to find derivative:

    [tex]f(x)'=\frac{(\sqrt[3]{x-4})' (x-1)-(\sqrt[3]{x-4})(x-1)'}{(x-1)^2}= [/tex] [tex]\dfrac{\dfrac{1}{3} (x-4)^{\frac{-2}{3}} (x-4)'(x-1) - (x-4)^{\frac{1}{3}}}{(x-1)^2}=[/tex] [tex]\dfrac{\dfrac{1}{3} (x-4)^{\frac{-2}{3}}(x-1) - (x-4)^{\frac{1}{3}}}{(x-1)^2}[/tex]

    2)Critical points are:

    a) 2 and 12

    b) [tex] \neg f'(x)[/tex] if x=1

    c) [tex]\dfrac{\dfrac{1}{3} (x-4)^{\frac{-2}{3}}(x-1) - (x-4)^{\frac{1}{3}}}{(x-1)^2}=0[/tex]

    here I stopped
     
  2. jcsd
  3. May 9, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are missing a critical point. You have to keep simplifying that expression to see it. Try multiplying numerator and denominator by (x-4)^(2/3).
     
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