# What does this symbol mean?

• Feb 6th 2009, 12:13 AM
champrock
What does this symbol mean?
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
• Feb 6th 2009, 03:45 AM
mr fantastic
Quote:

Originally Posted by champrock
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
• Feb 6th 2009, 06:19 AM
champrock
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
• Feb 6th 2009, 07:13 AM
HallsofIvy
Quote:

Originally Posted by champrock
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! $f(x)= x|x|e^{-x}$ so $f(x)= x^2e^{-x}$ for $x\ge 0$, $f(x)= -x^2e^{-x}$ for x< 0. In particular, for h positive, what is
$\lim_{h\rightarrow 0^+}\frac{f(0+h)- f(0)}{h}= \lim_{h\rightarrow 0^+}\frac{h^2}{h}e^{-h}$?

For h negative, what is
$\lim_{h\rightarrow 0^-}\frac{f(0+h)- f(0)}{h}= \lim_{h\rightarrow 0^-}\frac{-h^2}{h}e^{-h}$?
• Feb 6th 2009, 07:40 AM
champrock
hmm ur proof definately means that it is not continous. but i got this diagram drawn up from mathematica
http://img27.imageshack.us/img27/770...90901pmjx8.png

seems continous to me?
• Feb 8th 2009, 02:25 AM
champrock
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.)

http://img160.imageshack.us/img160/2...34207pmok5.png

http://img132.imageshack.us/img132/6...34207pmkh1.png
• Feb 8th 2009, 07:51 AM
HallsofIvy
Quote:

Originally Posted by champrock
hmm ur proof definately means that it is not continous. but i got this diagram drawn up from mathematica
http://img27.imageshack.us/img27/770...90901pmjx8.png

seems continous to me?

If you are referring to $f(x)= x|x|e^{-x}$, absolutely NOT. That is both continuous AND differentiable everywhere since both of the limits I showed are 0!
• Feb 8th 2009, 09:16 PM
champrock
Hi HallsofIvy

Actually they were both different questions. I am slightly confused. http://img222.imageshack.us/img222/4...34207pmhk9.png
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
• Feb 9th 2009, 09:31 AM
champrock
any opinions?
• Feb 9th 2009, 01:25 PM
HallsofIvy
Quote:

Originally Posted by champrock
Hi HallsofIvy

Actually they were both different questions. I am slightly confused. http://img222.imageshack.us/img222/4...34207pmhk9.png
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

No, no, no! the limit with h positive is $\frac{h^2}h e^{-h}= he^{-h}$ 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.

Quote:

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?
The two limits are both 0.

Quote:

(I know something is wrong in the above reasoning, can you please clarify it?)

Thanks
• Feb 10th 2009, 07:01 AM
champrock
well, if the two limits (RHL and LHL) are 0 for both the questions then both the functions are continous for all X? ? Is there anything wrong in this?