Question

I'm thinking about sketching functions of this sort
\(\huge \rm f(x)=xsin(1/x)\) or any function that involves sin or cosine of this 1/x that oscillates when it is near zero. My concern is to draw it in it's domain.

Answer 1

when i learn about sin(1/x) I know that it is oscillating near zero, after knowing useful tool of differentiation how could i draw this kind of functions from scratch

Answer 2

@perl some suggestions hear please

Answer 3

here*

Answer 4

@gorv

Answer 5

@ganeshie8

Answer 6

@iambatman mister batman can you have some suggestions here lol

Answer 7

Is this a general question on how to sketch the curve, \(f(x)=x\sin\left(\dfrac{1}{x}\right)\)?

Answer 8

yeah like how to sketch it using derivatives first and second
I actually have a hard time to sketch this kind of functions

Answer 9

my concerns it that i know how it looks like but can't sketch from scratch using what i learned from calc

Answer 10

Generally, to sketch a curve somewhat accurately, you need (1) the domain and range; (2) the intercepts; (3) critical points; (4) inflection points; and (5) asymptotes (both vertical and horizontal/oblique), if any of these exist.

Domain: all real \(x\), except for \(x=0\).
Range: all real numbers greater than the global minimum (which you can determine with the derivative test).

Your function doesn't have any \(y\) intercepts, because any point on the y-axis would have an x-coordinate of \(0\), which is not in the domain. It has infinitely many \(x\) intercepts, and you should be able to characterize them using some sort of pattern.

Critical points are easy enough to find (at least numerically). Inflection points will involve the same procedure.

There aren't any vertical asymptotes to worry about, I don't think... I'm not sure whether the singularity at \(x=0\) qualifies as an asymptote or not. Horizontal ones are much simpler to find, as you would just take the limits as \(x\to\pm\infty\).

Answer 11

yes i did the domain
the first derivative is not easy to deal with sinx-cosx/x
x intercept i think I'm fine with them.
you said critical points are easy to find numerically?
for horizontal asymptotes i get y=1 (from both infinities)

Answer 12

how would i find critical point solving sinx-cosx/x=0
one critical point is zero of course the rest, I was thinking about newtons method lol
but not sure

Answer 13

By "easy" I meant possible :P

Answer 14

okay! do you think i need newtons method
which i doubt will work since we so many zeros that have some pattern?

Answer 15

Yes, I but I wouldn't do it by hand. I'd suggest writing an algorithm to do it for you, if you have the know-how.

Answer 16

algorithm? interesting! can you show me how to do that

Answer 17

I'm kinda obsessed about this sort of functions lol
i tried to come up with some similar functions as well like xroot{1+sin(1/x)} and such

Answer 18

It would depend on the software you'd want to use. I have something saved as a Mathematica file (which can be opened using what's called a CDF reader, but I don't know if it's possible to run any commands) as well as a code written for the software R.

Answer 19

eh sounds too much work! but i'm willing to learn about this stuff hehe
please send me the file, I will try to install CDF reader and see. if I'm stuck i will bother you again lol. Are you a programmer as well?

Answer 20

Not really, no, but I had to learn a bit of the Mathematica language as part of my job, and what little code I've written in R was for a class.

Attached you'll find the Mma notebook, which I did not write (one of my professors did quite some time ago in one of the older versions), so all credit goes to him.

Answer 21

eh i see thank a lot, you have always been such a significant help^_^
my knowledge of programming doesn't exceed C++, but even that i forgot now lol

Answer 22

Fortunately, Mma is very similar to C++, so you shouldn't have too much trouble following along. If you run into any trouble with certain commands, you can search for info about them here: http://reference.wolfram.com/language/

Answer 23

*or so I've been told

Answer 24

that's good! i will try and see how that works

Answer 25

thanks again^_^

Answer 26

You're welcome!