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

kingstrick
Homework Statement

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