Question

can anyone help me look at the following question?
Please find attached. Thanks

Answer 1

jeez i can't

Answer 2

Actually, I would like to discuss this question with you. Can?
For this piecewise function to be differentiable, it must be continuous on the domain of R

Answer 3

if you take the derivative of the top expression you get
\[2x\cos(\frac{1}{x})+\sin(\frac{1}{x})\] if i am not mistaken
that oscillates wildly when you are close to zero, so i can't imagine how you are going to make the derivative exist at 0

Answer 4

And for the derivative at x = 0 and x=pi exist, the limit of the derivative must exist as well....

Answer 5

yes, so you really have two conditions
continuity and the derivatives must meet up as well

Answer 6

yah......

Answer 7

just want to check and confirm if my a,b,c,d are correct values or not

Answer 8

contiuity at x = 0 and pi exist. the left hand continuity = the right hand continuity

Answer 9

well we have with continuity two things
\[d=0\] from the first one and
\[a\pi^2+b\pi^2+c\pi=0\]

Answer 10

oops i meant \(a\pi^3+b\pi^2+c\pi=0\)

Answer 11

yes. i agree. I get c = 0 as well

Answer 12

from the derivative?

Answer 13

from the limit of derivative

Answer 14

\[2a\pi^2+2b\pi+c=1\] is what i get

Answer 15

well except that it is wrong, should be \(-1\)

Answer 16

don't you just evaluate the piecewise function?

Answer 17

you lost me

Answer 18

use the above formula. but for this limit to exist, left hand limit must be equal to right hand limit

Answer 19

when x approaches pi and 0

Answer 20

that is the fancy way to do it
the derivative of sine is cosine, \(\cos(\pi)=-1\) so you have to get \(-1\) from the middle one

Answer 21

but i am still trying to figure out what you are going to do with
\[2x\cos(\frac{1}{x})+\sin(\frac{1}{x})\] because there is no way i can see to make this work at \(0\)

Answer 22

you can use the squeeze theorem

Answer 23

for the case of cos(1/x). for cos funciton, cos is always bound to [-1.1]. so 2x*cos(1/x) is bound to [-2x.2x]

Answer 24

that part is not the problem

Answer 25

the limit of 2x and -2x when x goes to 0 is 0, so by squeeze theorem, the limit of 2x*cos(1/x) when x approaches 0 is 0 as well

Answer 26

at 0 no limit exists

Answer 27

that is the problem

Answer 28

it exists. by squeeze theorem. you may check it out . squeeze theorem

Answer 29

derivative oscillates infinitely often between \(-1\) and \(1\) as you approach 0 from the left

Answer 30

http://www.sosmath.com/calculus/limcon/limcon03/limcon03.html

Answer 31

what you are saying is that
\[\lim_{x\to 0^-}x^2\cos(\frac{1}{x})=0\] and i certainly agree with that

Answer 32

okay. so what is next. still havent find the values of a and b

Answer 33

no theorem necessary as cosine is stuck between -1 and 1
what i am saying is, this beast is not differentiable at \(x=0\)

Answer 34

if it was you would match up the derivatives and see what you get
but it isn't
i can't think of any way you are going to make it differentiable at 0

Answer 35

if at x = 0, the attached formula. the left hand limit == right hand limit. then f is differentiable at 0. that is why we need to find the values of a and b

Answer 36

but your problem is you cannot compute that limit, because \(f(0)\) does not exist

Answer 37

you can define f(0) to be 0

Answer 38

lets look at the picture
http://www.wolframalpha.com/input/?i=2x*cos%281%2Fx%29%2Bsin%281%2Fx%29+domain+-.2..0

Answer 39

anyway, I need someone to help me solve the problem

Answer 40

ok @nincompoop will get it i am sure

Answer 41

thanks!

Answer 42

better yet - http://www.wolframalpha.com/input/?i=limit&a=*C.limit-_*Calculator.dflt-&f2=2x*cos(1%2Fx)%2Bsin(1%2Fx)&f=Limit.limitfunction_2x*cos(1%2Fx)%2Bsin(1%2Fx)&f3=0&f=Limit.limit%5Cu005f0&a=*FVarOpt.1-_**-.***Limit.limitvariable--.**Limit.direction--.**Limit.limitvariable2-.*Limit.limit2-.*Limit.direction2---.*--

Answer 43

I am getting the hang of using wolframalpha lol. but my school uses maple, so I start getting the syntax of the two mixed up

Answer 44

this is a piecewise function. so you shouldn't just look at one single function only