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Critical points

  • Thread starter Lynne
  • Start date
  • #1
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Homework Statement


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]

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
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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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