Question

Someone help me understand the solution, please
http://mathforum.org/library/drmath/view/51951.html

Answer 1

What about it doesn't make sense?

Answer 2

we need prove both ways
1) suppose |f(x) | is diff at x =c, then f'(c) =0
2) if f'(c) =0, then |f(x) | is diff at x =c
right?

Answer 3

I think tha the condition f '(c)=0 is a too much strong condition

Answer 4

1) if |f(x)| is diff at x =c, then |f(c)|' ( not |f' (c)|) exist and it is ....
it doesn't relate to f'(c)

Answer 5

f is a diff. function at everywhere, hence f is diff at c and
\(f'(c) =\lim_{x\rightarrow c}\dfrac{f(x)-f(c)}{x-c}\)
Assume \(|f(x)|\) is diff at c, then
\(|f(c)|' =\lim_{x\rightarrow c}\dfrac{|f(x)|-|f(c)|}{x-c}\)

Answer 6

since f(c) =0 hence |f(c)| =0 , we have each of them have different value,
they don't link together. :(

Answer 7

@eliassaab

Answer 8

Well to be honest I didn't read the solution. You didn't like it anyway and I find all this 'ascii art' really annoying...

So, the question was:
"Suppose that \(f: R \to R \)is differentiable at c and that \(f(c) = 0\).
Show that \(g(x) = |f(x)|\) is differentiable at c if and only if \(f'(c) = 0\)."

Let's see, why would we care what if the value of \(f(c)\) is 0? Well, that is because what the absolute value does, it will make anything negative into positive and basically 'flips that up'.
Now, suppose \(f'(c) \neq 0\), that means the function is 'raising' or 'falling' at that point.
So for \(f'(c) < 0\), it means the function is positive on the left of c and negative on the right. (in some neighborhood around c of course).
\(f'(c) > 0\) means just the opposite, same idea.

But here is the thing, in order for \(f(x)\) to be differential at c the limit of the derivative has to give the same value no matter from which direction \(\Delta x\) approaches 0:
$$
f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}
$$
If \(f'(c) \neq 0\), then by operating the absolute value you 'flip' the negative part to be positive and make a function that instantaneously stop falling and start raising at that point.

That has the shares the problem that the absolute value function itself has at x=0, you get different values for the derivative limit at that point because of that 'flip' and therefore it is not differentiable there.

Answer 9

@pitamar see this
http://math.stackexchange.com/questions/807537/function-which-derivative-at-0-is-1-but-is-not-monotonic-increasing

Answer 10

The solution
http://mathforum.org/library/drmath/view/51951.html
Is easy to read and understand and seems correct to me.

Answer 11

@eliassaab Noted. I used the misleading words when saying 'growing' or 'falling'. However, I have meant to the function's graph.
So when I said 'growing' I meant that in some neighborhood, the points' values on the left are smaller and on the right are greater.

The solution doesn't state much better, it says:
"If you look at a function near a point where it crosses the axis, and
if the function has a non-zero slope there (so f'(c) =/= 0, where
=/= means "is not equal to"), then the graph will look like
\ /
/ \ /
------/----- , so g(x) = |f(x)| looks like ---------- ,
/
because taking the absolute value reflects or folds the part of the
graph below the x-axis up into the part above the axis. That
intuitively explains why f'(c) must = 0 in order for |f(x)| to be
differentiable at c. You should be able to see that, in that case,
the derivative of g at c, g'(c), should also be 0, so we expect L=0."

And goes on by showing that positive derivative ought to mean that because of the definition of the derivative and that the 'folding' will flip the sign for one side of the limit.
I was trying to say the same thing..

I feel that I have used the same argument, but if you feel it is unclear or inaccurate.. Well, I can't blame you =)

Answer 12

Here attached is an explanation of what is going on
@ganeshie8 @pitamar @Loser66

Answer 13

@eliassaab That is a nice solution, but I still think that the argument is the same. One way or another they all say exactly the same thing =|

Answer 14

Agree 100% pitamar

Answer 15

I found out the other way to prove it.
suppose f'(c) =0 , \(f'(c) = lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}=0\)
that is for \(|x-c|<\delta\)
\(|\dfrac{f(c+h)-f(c)}{h}|<\varepsilon\)
but the left hand side
\(|\dfrac{f(c+h)|-|f(c)|}{h}\leq|\dfrac{f(c+h)-f(c)}{h}|<\varepsilon\)
the far left is lim of |f(x)| when x --> c, that shows that limit exists and hence |f(x)| differentiable.

Answer 16

Do not forget that f(c)=0 . You also need to prove the other direction.