Question

a)Suppose g:R--> R is an everywhere differentiable function. Show that if g(a) \(\neq 0\) then |g(x)| is differentiable at x =a. Hint: use the fact that composition of 2 differentiable functions is differentiable.
b) Suppose g(a) =0. Show that |g(x)| is differentiable at x =a if and only if g'(a) =0
Please, help

Answer 1

@kmeis002

Answer 2

Ah, with the hint, note that for the function \(g = abs(x) = |x|\) is differentiable everywhere except \(x=0\).
From the hint, we know that if \(f(x), g(x)\) are differentiable then \(f(g(x))\) is differentiable. Since \(|x|, f(x)\) are differentiable everywhere but 0 implies that \(|f(x)|\) is differentiable there.

Answer 3

Thinking about b briefly

Answer 4

How f(|x|) = |f(x)|?

Answer 5

Ah, we are not claiming that \(f(|x|) = |f(x)|\). Just that if you compose two differentiable functions, the new function is differentiable. So something like
\[h(x) = f(g(x))\]
If \(f(x), g(x)\) is differentiable at a than \(h(x)\) is differentiable there. In this case we let \(f(x) = |x|\) and \(g(x) = g(x)\). Note that \(|x|\) is differentiable everywhere but 0

Answer 6

Yes, it makes more sense. Thanks for that part

Answer 7

I need to head to bed soon, but the general idea of the argument is this;
If \(g'(a) \neq 0\) at \(a=0\) then \(g(x)\) is changing signs at \(x=0\). For the sake of argument, lets assume \(g(0+h) > 0\) and \(g(0-h) < 0\). Then if we look at the definition of the derivative from left/right limits:
\[\lim_{h \to 0^+} \frac{g(0+h)-g(0)}{h} = \lim_{h \to 0^+} \frac{g(0+h)}{h} > 0 \]
and since h < 0 when approacing from the left, we get
\[\lim_{h \to 0^-} \frac{g(0+h)-g(0)}{h} = \lim_{h \to 0^-} \frac{g(0+h)}{h} > 0 \]
Note that in this case:
\[\lim_{h \to 0^-} \frac{g(0+h)-g(0)}{h} = \lim_{h \to 0^+} \frac{g(0+h)-g(0)}{h} \]
and so the derivative \(g'(0)\) exists.

When we take the absolute value of \(g(x)\) we now get \(|g(0+h)| > 0\) and \(|g(0-h)| > 0\). Since both limits now have a positive numerators, we get that the limits cannot equal each other and so the limit doesn't exist:
\[\lim_{h \to 0^+} \frac{|g(0+h)|-|g(0)|}{h} = \lim_{h \to 0^+} \frac{|g(0+h)|}{h} > 0 \]
and since h < 0 when approacing from the left, we get
\[\lim_{h \to 0^-} \frac{|g(0+h)|-|g(0)|}{h} = \lim_{h \to 0^-} \frac{|g(0+h)|}{h} < 0 \]
Since one is positive and the other is negative, we get:
\[\lim_{h \to 0^-} \frac{|g(0+h)|)}{h} \neq \lim_{h \to 0^+} \frac{|g(0+h)|}{h} = DNE \]

When \(g'(0) = 0\) and \(g(0)=0\), the function \(\g(x)\) retains it sign and so the absolute value of \(|g(x)|\) will exist.

Answer 8

Thanks a lot, good night though. It is so late here also. :)

Answer 9

No problem. Just as note, if \(g(x)\) is strictly \(g(x) > 0 \) or \(g(x) < 0\) implies that \(|g(x)| = g(x)\) or \(|g(x)| = -g(x)\), both of which are differentiable everywhere. The contradiction only arises when it changes sign at 0.

Glad I could be of some help.

Answer 10

Not some, you give me a ton of help. :)

Answer 11

My post above should just be, "when it changes sign", to be the most general.