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!

Show that g(x):=|f(x)| is differentiable at c

  1. Apr 10, 2012 #1
    The problem statement, all variables and given/known data

    Suppose that f:ℝ→ℝ is differentiable at c and that f(c) =0. Show that g(x):=|f(x)| is differentiable at c iff f'(c)=0.


    The attempt at a solution

    **This is the solution that I was shown by a peer but I do not understand it... Can anyone break it down for me. It feels like some steps were skipped or taken out. I just don't follow it.

    Proof:

    Suppose that f:ℝ→ℝ is differentiable at c and that f(c) =0. Then (g(c+h)-g(c)) / h = (|f(c+h)|-|f(c)|)/h = |f(c+h)|/h. Now suppose f'(c)=0 then as h→0, |f(c+h)|/h=|f(c+h)-f(c)|/h →|f'(c)|=0. Therefore -|f(c+h)|/h ≤ |f(c+h)|/h ≤ |f(c+h)|/h, as h→0, |f(c+h)|/h→0, then lim (as h→0) (g(c+h)-g(c))/h=0. Therefore g'(c)=0 and g is differentiable at c.

    Now suppose g(x) is diff at c. →lim (as h→0) (g(c+h)-g(c))/h exists so lim (as h→0) |f(c+h)-f(c)|/h = lim (as h→0) |f(c+h)|/h exists <--> lim (as h→0+)||f(c+h)|/h and lim(h→0-)|f(c+h)|/h both exists and are equal. but lim (h→0+) |f(c+h)|/h≥0 while lim(h→0-) |f(c+h)|/h≤0, hence lim (h→0+) |f(c+h)|/h = lim(h→0-) |f(c+h)|/h =lim(h→0) |f(c+h)|/h = 0 → |f(c+h)|/h = 0 since -|f(c+h)|/h ≤ (f(c+h)-f(c))/h=f(c+h)/h ≤|f(c+h)|/h. Therefore lim (as h→0) ((c+h)-f(h))/h=lim (as h→0) f(c+h)/h. Hence f'(c)=0.

    Therefore g(x):=|f(x)| is diff. at c iff f'(c)=0



    I am totally lost....please help.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?