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Absolute max and absolute min problem

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the absolute maximum and absolute minimum values of f(x) = (x - 1)^2/3 on [0,2]. Give reasons for your answer

    2. Relevant equations
    None


    3. The attempt at a solution

    I first differentiated the function and got. [tex]\frac{2}{3}[/tex]x(x - 1)[tex]^{\frac{-1}{3}}[/tex]

    that is to the power of -1/3, not multiplied by.

    Then I set that equal to 0, and solved for x, this is where I think I got something going wrong. I got x = 0 as the critical point. If I set it equal to 2, then I get either x = 0 or x = -54, which is of course wrong.

    What should I do
     
  2. jcsd
  3. Oct 7, 2007 #2
    how did you get x(x-1)? shouldn't it just be 2/3*(x-1)^(-1/3)?
     
  4. Oct 7, 2007 #3
    Chain rule. You differentiate what is inside the brackets too and then multiply that
     
  5. Oct 7, 2007 #4
    yea but d(x-1)/dx=1 o.o

    so if you do it the long way with a substitution:

    u=x-1

    [tex]f(u)=u^{\frac{2}{3}}[/tex]
    [tex]\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}[/tex]
    [tex]=\frac{2}{3}u^{\frac{-1}{3}} * 1=\frac{2}{3}(x-1)^{\frac{-1}{3}}[/tex]
     
    Last edited: Oct 7, 2007
  6. Oct 7, 2007 #5
    ****
     
  7. Oct 7, 2007 #6
    So now we can't find x when f ' (x) = 0 ?
     
  8. Oct 7, 2007 #7
    Yea I get it, I just didn't think too well when I first differentiating it. Thanks for pointing that out though.
     
  9. Oct 7, 2007 #8
    you can still find x when f'(x)=0 just that there's only 1 possible option for x.
     
  10. Oct 7, 2007 #9
    I got, x = 35/27 as the absolute minimum and x = 0 and 2 as the maximum, this doesn't sound right
     
    Last edited: Oct 7, 2007
  11. Oct 7, 2007 #10
    how did you get 35/27? and yes check the endpoints are they a max/min?

    and just a conceptual question, why can't 1 be a max/min? if you take the limit as x -> 1 it gives a large number but can it be a max/min?
     
  12. Oct 7, 2007 #11
    Actually 1 is probably the absolute minimum. not 35/27

    I got 35/27 by setting f ' (x) = 1 and therefore

    1 = 8/(27x - 27)

    x = 35/27
     
  13. Oct 7, 2007 #12
    no f'(x)=0 will give you the critical points, but you also consider the end points iff the interval given is closed which is in this case.

    also the critical points must be in the domain of f(x) so 1 can't be it. so the endpoints will most likely be the max/min
     
  14. Oct 7, 2007 #13

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The maximum and minimum values on a closed and bounded interval (if the interval is not closed or not bounded the function may not have max or min) occur at three kinds of points: in the interior where the derivative is 0, at an end point, in the interior where the derivative does not exist.

    As Bob1182006 said, f'(x)= (2/3)(x-1)-1/3 which is never 0. It is, however, undefined at x= 1. The only possible x values for the max and min are x= 0, x= 1, and x= 2. Put those into the function f(x) and see what you get.

    Bob1182006, 1 is in the domain of f(x). It just isn't in the domain of f'(x).
     
  15. Oct 7, 2007 #14
    It sometimes help to draw a horizontal line putting all the numbers X must be so f'(x) = 0 (or undefined). Then you put some numbers in f(x) like 0 or 2 (numbers on both sides of 1) and see what happens. If you get negative number, the function (f(x)) is decreasing, if it is a positive number the function is increasing

    In your case it will look like this.

    ................................ 1
    ________________________________________
    f'(x) - - - - - - - - - - - 0 + + + + + + + + + +

    To find the on and only critical point, put 1 into the f(x) equation and see what you get.

    Hence, the CP is (x, f(1))

    Edit: Don't mind the dots I typed, I couldn't get the "1" to stick right above the zero...
     
    Last edited: Oct 7, 2007
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