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I must go to the higher derivatives?

  1. Apr 7, 2014 #1
    Function ##f(x)=x^4## has minimum at ##x=0##.
    ## f'(x)=4x^3##
    ##f'(0)=0##
    ##f''(x)=12x^2##
    ##f''(0)=0##
    ##f^{(3)}(x)=24x##
    ##f^{(3)}(0)=0##
    ##f^{(4)}(0)>0##
    So what is the rule? I must go to the higher derivatives if ##f'(0)=f''(0)=0##?
     
  2. jcsd
  3. Apr 7, 2014 #2

    Mark44

    Staff: Mentor

    No. You have f'(0) = 0 and f' changes from negative to positive as x changes, going left to right through zero. That's enough for you to conclude that there is at least a local minimum at zero.

    Also, since f''(x) = 12x2, which is positive everywhere except at x = 0, the graph of f is concave upward, so again there is a minimum at x = 0.
     
  4. Apr 7, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    As Mark44 said, you can use the "first derivative test". But what you are saying is true. You can determine whether f(x) has a max or min by checking the first non-zero derivative at that point.

    To see that think about writing the Taylor's polynomial for f up to that first non- zero derivative:
    [tex]f(a)+ f'(a)(x- a)+ \frac{f''(a)}{2}(x- a)^2+ \cdot\cdot\cdot+ \frac{f^{(n)}}{n!}(x- a)^n[/tex]
    [tex]= f(a)+ \frac{f^{n}(a)}{n!}(x- a)^n[/tex]
    If n is odd x= a is a saddle point. If n is even, it is either a maximum or a minimum. If [itex]f^n(a)> 0[/itex] f increases away from x= a so that is a minimum. If [itex]f^n(a)< 0[/itex], it decreases away from x= a so that is a maximum.
     
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