1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Max and Min Values

  1. Nov 16, 2005 #1
    OK, the professor did this problem for us as an example and i got lost somewhere in between. the problem was: find the critical points
    f(x)= x^(4/5)*(x-4)^2

    then he used the product rule to get

    f '(x)= 4/5 x^(-1/5) *(x-4)^2 + x^(4/5) * 2(x-4) = 0

    THEN, the part that threw me off was the next part where he said multiply both sides by 1/5....what did he mean by that? after that you get

    4/5 (x-4)^2 + 2x(x-4) = 0 and so on...

    my question is how and what did he do to get rid of the x^(-1/5) and the x^(4/5)??? :confused:

    Any help would be greatly appreciated
  2. jcsd
  3. Nov 16, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Multiply both sides by
  4. Nov 16, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    He multiplied both sides by [tex]x^{1/5}[/tex], not 1/5.
  5. Nov 16, 2005 #4
    Don't multiply by 1/5, try multiplying by x^(1/5). The whole idea is to turn the exponents into integers, and then you can use the usual methods to find the roots of the equation.
    Last edited: Nov 16, 2005
  6. Nov 16, 2005 #5
    He is not multiplying by 1/5, he is multiplying by x^(1/5). On the zero side, it of course goes away. On the other side, it distributes and x^(1/5)*x^(-1/5)=1 while x^(1/5)*x^(4/5)=x nicely killing off those pesky rational exponents. Good Luck
  7. Nov 16, 2005 #6
    WOOHOOO!!!!! Thanks very much!
  8. Nov 16, 2005 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    "Never seen such unanimity of opinion before in my life"
    Poobah, in "The Mikado"
  9. Nov 17, 2005 #8
    And who said mathmaticians don't have a sense of humor??

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Max and Min Values