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First derivative test

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the equation below. (Round the answers to two decimal places. If you need to use -infinity or infinity, enter -INFINITY or INFINITY.)
    f(x) = x^4 - 50x^2 + 4

    Find the local minimum and maximum values of f.


    2. Relevant equations

    First derivative test and that is about it


    3. The attempt at a solution

    Take the derivative using the power rule so
    f ' (x)= 4x^3-100x

    Then to find critical points, set f' (x)=0 and solve for x

    0=4x^3-100x
    0=4x(x^2-25), x=-5,0,5

    then plug those x values into the original function and you get

    f(0)=4 ------max
    f(-5)= -621-----min
    f(5)=-621------min

    This is what i get for an answer, however the where I can type these answers online there are only two boxes. One is for min and one is for max. What did i do wrong?
     
  2. jcsd
  3. Nov 15, 2009 #2
    How are you determining those to be local maximums and minimums just based upon their value? I think you are confusing the first derivative test for local extrema and the first derivative test for absolute extrema on closed intervals. You are dealing with the entire real line and asked to find the local extrema.

    Once you have found your critical points, that is x=-5,0,5, you break up the real line into intervals where you exclude these points. So you will get (-∞,-5), (-5,0), (0,5), and (5,∞). Then choose a point in each interval and evaluate f'(x) at that point. If f' is positive at that point, then the function f is increasing on that interval, if it is negative, then f is decreasing on that interval.

    If to the left of a critical point, the function is increasing, and to the right, the function is decreasing, then that point gives you a local maximum. You carry on in this way, until you have categorized each critical point as giving either a local maximum, minimum, or neither. Evaluate f at those points to actually get the maximum/minimum values.
     
  4. Nov 15, 2009 #3
    But isn't that what i did?
     
  5. Nov 15, 2009 #4
    There are three critical points. By seeing what intervals of the four go up and down, I believe you get the same answer i do. No?
     
  6. Nov 15, 2009 #5
    No it isn't. Did you read what I wrote? You just evaluated f at the critical points, and that was it. You got lucky that in this problem all three critical points gave extrema, but that isn't always going to be the case. For example, look at the following http://www.wolframalpha.com/input/?i=plot+x^4-2x^3-2". Notice that f'(x)=0 at x=0, but it is neither a maximum nor a minimum.

    By your method above, you would have gotten that x=0 gave a maximum value simply because it is larger than the value of the function evaluated at the other critical point.
     
    Last edited by a moderator: Apr 24, 2017
  7. Nov 15, 2009 #6
    I still get the same answer using your method. Intervals that are increasing include (-5,0),(5,infinity) and decreasing intervals are (-infinity,-5),(0,5).

    "If to the left of a critical point, the function is increasing, and to the right, the function is decreasing, then that point gives you a local maximum. You carry on in this way, until you have categorized each critical point as giving either a local maximum, minimum, or neither. Evaluate f at those points to actually get the maximum/minimum values."

    going through it decreases on -infinity to -5 and then increases from - 5 to 0. local min. -5 to 0 it increases and 0 to 5 it decreases. local max. 0 to 5 it decreases and 5 to infinity it increases. local min.
     
  8. Nov 15, 2009 #7
    i still dont understand what im doing wrong so
     
  9. Nov 15, 2009 #8
    nevermind. i figured it out.
     
  10. Nov 15, 2009 #9
    Yes, you got the same answer because you got lucky. You're method was just wrong. Also, the answer only has two answer boxes because they are asking for the maximum and minimum values.
     
  11. Nov 16, 2009 #10
    I understand now. I learned the correct way from you on how to do it. The value thing tripped me up. thanks for the help.
     
  12. Nov 16, 2009 #11
    No problem!
     
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