1. Limited time only! Sign up for a free 30min personal 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!

Relative extrema of|f(x)|

  1. Jan 24, 2010 #1
    Use any method to find the relative extrema of [tex]|1+\sqrt[3]{x}|[/tex].
    I get the following [tex] |1+\sqrt[3]{x}| = \left\{\begin{array}{cc} 1+\sqrt[3]{x}, & \mbox{if} x >-1 \\ -1-\sqrt[3]{x}, & \mbox{if} x<-1\end{array}\right[/tex]
    Using the first derivative test, I get [tex] f'(x) = \frac{1}{3x^{2/3}} \mbox{and} -\frac{1}{3x^{2/3}}[/tex]. These suggest that there is a relative extremum at x=0 but not at x=-1, actually it is an inflection point at x=0. What is next? Maybe I am confused about absolute values! I think I have the function piece wise ok.Please help. Thanks.
     
  2. jcsd
  3. Jan 24, 2010 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    At x=0, the derivative is infinite. I am confused about what you are confused about.
     
  4. Jan 25, 2010 #3
    Should not f'(x) =0 show a relative minimum at x=-1, i.e., if f(x) is differentiable at x=-1. According to my graphing calculator there is a relative minimum at x=-1. But I do not know to find the relative minimum at x=-1 by calculus? Thanks.
     
  5. Jan 25, 2010 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    f(x) is not differentiable at x=-1.

    That can be seen by the different values the right-hand derivative and left-hand derivative gets there.

    Thus, x=-1 is a sharp corner, and it is, indeed, a relative extremum.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relative extrema of|f(x)|
  1. Proof f'(x)/f(x)=|f(x)| (Replies: 26)

  2. Solve f(-x)=-f(x)? (Replies: 7)

  3. Integrating f'(x)/f(x) (Replies: 8)

  4. Ln(x) =f(x) (Replies: 2)

Loading...