HI
Can anyone please tell me what is meant by that symbol between X and X ? I can do the questions myself but I dont know how to interprete that?
http://img23.imageshack.us/img23/8/26200914210pmzs8.png
Thanks
HI
Can anyone please tell me what is meant by that symbol between X and X ? I can do the questions myself but I dont know how to interprete that?
http://img23.imageshack.us/img23/8/26200914210pmzs8.png
Thanks
Read this: Elementary Calculus: Absolute Value Function
oh ya i know absolute function. somehow i got confused and thought it was something else. just for confirmation , the answer is B right? it is a continous function everywhere and differentiable everywhere. But it is not differentiable at x=0?
Thanks a lot
I suiggest you look more closely! $\displaystyle f(x)= x|x|e^{-x}$ so $\displaystyle f(x)= x^2e^{-x}$ for $\displaystyle x\ge 0$, $\displaystyle f(x)= -x^2e^{-x}$ for x< 0. In particular, for h positive, what is
$\displaystyle \lim_{h\rightarrow 0^+}\frac{f(0+h)- f(0)}{h}= \lim_{h\rightarrow 0^+}\frac{h^2}{h}e^{-h}$?
For h negative, what is
$\displaystyle \lim_{h\rightarrow 0^-}\frac{f(0+h)- f(0)}{h}= \lim_{h\rightarrow 0^-}\frac{-h^2}{h}e^{-h}$?
i studied on continuity and thanks for pointing out that mistake of mine.
But , I am again stuck. Going by the logic of the previous question, the following function should also be:
1. Continous everywhere but Not continous at x=0
2. Differentiable everywhere except x=0
But this is not an option at all in the question. can you please guide me on this? (Again, I am not asking for an answer, but just help which can probably help me figure out the thing myself.)
Hi HallsofIvy
Actually they were both different questions. I am slightly confused.
In the first question, the right hand limit was positive (+h) and the left hand side limit was negative (-h). That is why the function was not continous at x=0 because RHL will not be equal to LHL
Following the same reasoning, in the second question, the right hand limit is positive (h*e^-h) and the left hand limit is negative (-h*e^-h). {since e^-h is positive always) So, the RHL is not equal to LHL in this case also. So, it is not continous?
(I know something is wrong in the above reasoning, can you please clarify it?)
Thanks
No, no, no! the limit with h positive is $\displaystyle \frac{h^2}h e^{-h}= he^{-h}$ and that goes to 0(1)= 0. The limit with h negative is [tex]\frac{h^2}{-h}e^{-h}= -he^{-h} and that goes to -(0)(1)= 0.
The two limits are both 0.Following the same reasoning, in the second question, the right hand limit is positive (h*e^-h) and the left hand limit is negative (-h*e^-h). {since e^-h is positive always) So, the RHL is not equal to LHL in this case also. So, it is not continous?
(I know something is wrong in the above reasoning, can you please clarify it?)
Thanks