# discontinuous derivatives

• Jun 8th 2010, 03:17 AM
raulk
discontinuous derivatives
Find a differentiable function g: R-R such that g′(0)=1>0, but g is not monotonic increasing in any interval (0, a).

The derivative is clearly not continuous on 0.
g′ (x) >0 for any x>0.
could anyone give me some hints how to come up with one

• Jun 8th 2010, 03:52 AM
First, try to write the derivative at zeros as: $\displaystyle f'(0)=\lim_{x->0}\frac{f(x)-f(0)}{x}$. Your objective is to have that limit equal to 1 and at the same time, $\displaystyle f'(t)\leq 0$ for $\displaystyle x>0$.
• Jun 8th 2010, 04:19 AM
raulk
I understand what you wrote. But I'm still not sure how to come up with an answer. I'm thinking about something with cos x. Could you please be a bit more specific? Thanks a lot for your help!
• Jun 8th 2010, 07:40 AM
drjerry
keep at it...
Mohammad's hint is a good one ;)

I'll submit the following clarification. To say that the function cannot be monotonically increasing on any positive interval (0, a), no matter how small $\displaystyle a$ is, does not mean that $\displaystyle f'(x)\leq 0$ has to be true for all $\displaystyle x > 0$. It suffices to have the function oscillate wildly in every arbitrarily small interval (0, a).

Can you think of a function that oscillates wildly? Now can you think of some factors by which to multiply so that the $\displaystyle x$'s in the formula Mohammad wrote down cancel out -- or cancel well enough so that the limit equals 1?

Cheers!
• Jun 8th 2010, 08:41 AM
hahahha...Thank you so much...You clarified the things for me too.... Now I have the exact answer. I was looking for a function that is decreasing after 0. That turned out to be wrong. I had the answer in front of me all the time. And raulk: you are close (Cool)
• Jun 8th 2010, 08:59 AM
raulk
g(x)=xcosx would meet the first requirement, but I guess g'(x)=cosx+xsinx isn't oscillating wildly enough.
I'm also trying to do something to the function f(x)=sin(1/x), but x cannot be zero.
Could you please give me some further hints how to fix this?
• Jun 8th 2010, 09:28 AM
drjerry
...that's why I said, to try "multiplying by some factor to get the $\displaystyle x$'s to cancel..."

You're really close, and you're right: $\displaystyle \sin(1/x)$ cannot be zero. But can you multiply that by something to "squash" the function down so that the limit exists as $\displaystyle x\to 0^+$ ? It's still going to oscillate wildly in all neighborhoods of the origin.

Look at the formula for the derivative that Mohammad if you need inspiration for factors... ;) And don't try to make the derivative $\displaystyle f'$ continuous at the origin! (As you've already noted.)

Hope this gets you there.
Jerry
• Jun 8th 2010, 09:37 AM
raulk
would g(x)=x^2 sin(1/x) work? I'm still not so sure.
• Jun 8th 2010, 10:18 AM
drjerry
I was thinking along different lines, but let's follow your lead... $\displaystyle x^2\sin(1/x)$ "works."

...but wait: you wanted $\displaystyle g'(0) = 1$, right? The function you proposed satisfies $\displaystyle g'(0) = 0$; at least $\displaystyle \lim_{x\to 0} g'(x) = 0$. Now what else could you do to your function to "raise" its derivative up to the desired value.

Things are starting to get tricky: the function you proposed oscillates wildly in all neighborhoods of the origin, but the oscillations remain between the curves $\displaystyle y = x^2$ and $\displaystyle y = -x^2$; draw a few graphs to convince yourself, please. If you want to add some function that makes the derivative satisfy $\displaystyle g'(0) = 0$, that function has to be "flat enough" that it also lies between these two curves. Otherwise the oscillation behavior is killed.

Think about adding functions whose graphs still lie between the curves $\displaystyle y = x^2$ and $\displaystyle y = -x^2$ close to the origin, and do so with the derivative at $\displaystyle x=0$ in mind. The right polynomial should do it. You don't have to worry about the derivative of $\displaystyle x^2\sin(1/x)$ at $\displaystyle x=0$ anymore; it's zero, and won't affect anything added to it.

Jerry