Question

Define f(x) ={\(x^2+2x^2sin\frac{1}{x}~~~x\not=0\);\(0~~x=0\)} Show that f'(0)>0.

Answer 1

@ganeshie8 , I am confused because the derivative will be 0 at 0, right?

Answer 2

by squeeze it is zero

Answer 3

derive and plug in zero for x into the derivative(?)

Answer 4

yea, I get 0 is the issue

Answer 5

zero for f'(x) (before plugging in zero for x into the derivative) ?!

Answer 6

I am sure you can derive x^2,
and the 2x^2sin(1/x) by the product rule, with a chain for sin(1/x).

Answer 7

omg... ok ignore me now.

Answer 8

give me a minute, i was still doing the last problem's limit

Answer 9

sure

Answer 10

maybe you need help on the last problem (whatever it is) ?

Answer 11

so for the deriv. I have 2x+4xsin(1/x)+-2cos(1/x) which gives me a dilemma at x=0

Answer 12

your derivative will not be defined at x=0.

Answer 13

but you are not getting the correct derivative.

Answer 14

?

Answer 15

\[f'(0) = \lim\limits_{x\to 0} \dfrac{f(x) - f(0)}{x-0} = \lim\limits_{x\to 0}~ x + 2x\sin(1/x) \]

Answer 16

f(x) = x^2+2x^2 sin(1/x)
f '(x) = 2x + 4x sin(1/x) + 2x^2 cos(1/x) * d/dx(1/x)

Answer 17

f '(x) = 2x + 4x sin(1/x) + 2x^2 cos(1/x) * (-1/x^2)

Answer 18

you want to apply squeeze thm for evaluating that limit ?
let me grab a paper in between

Answer 19

Chain for the inner function of the sine.

Answer 20

I fail to see the difference between our answers

Answer 21

but anyways, thanks ganesh, I think I'm gonna go from the def. Good idea

Answer 22

I will not disturb, but I'll only say that when you tried to derive the second function, the derivative of sin(1/x) is not just cos(1/x), it is cos(1/x) times the derivative of 1/x.
(CHAIN RULE)

Answer 23

which is indeed 0
http://www.wolframalpha.com/input/?i=+lim+%28+x+%2B+2x%5Csin%281%2Fx%29%29%2C+x-%3E0

Answer 24

@ganeshie8 , I'm getting 0 after squeeze though

Answer 25

so it doesn't really help in showing what u want to show.. hmm

Answer 26

Do your definition, as ganeshie says.

Answer 27

yes, and \(-1/x^2 *2x^2=-2\)

Answer 28

I simplified solomon

Answer 29

what ?

Answer 30

yea, that is the problem I am having. Then on top of that, part b is to say it is only increasing in any interval containing 0 which isn't true

Answer 31

my bad, icy...

Answer 32

you can't simply take derivative and evaluate the limit because we don't even know if f(x) is differentiable at x = 0. All we know is that i is continuous at x = 0. You're supposed to show f'(0) > 0. Start from @ganeshie8's equation

Answer 33

the issue sour is that the limit equals 0

Answer 34

I graphed it to take a look https://www.desmos.com/calculator

Answer 35

And within a neighborhood of 0, the deriv=0 like between .5,.5 or it is an itty bitty fluxuation between pos and neg

Answer 36

ok I have a type o

Answer 37

it's \(\color\red{x}+2x^2sin\frac{1}{x}\)

Answer 38

which changes ganeshi'es and makes it work. Thanks guys, sorry for my idiocy

Answer 39

ugh

Answer 40

:O

Answer 41

I had copied the question wrong on my paper. but at least the confusion was warranted

Answer 42

np :) i was thinking the same, making the firs term x gives leaves a constant term so that we get some number for derivative

Answer 43

yea, that's what I got, but now my issue is that part b is show that f is not increasing in any interval containing 0. But the whole function is increasing...?

Answer 44

and I just quadruple checked no type-o here

Answer 45

if f'>0 that implies increasing no?

Answer 46

any thoughts?

Answer 47

would you prefer I make a new question?

Answer 48

yes, we can try showing that the derivative oscillates between +ve andn -ve values as it approaches 0

Answer 49

http://www.wolframalpha.com/input/?i=%28x%2B2x%5E2sin%281%2Fx%29%29%27

Answer 50

I found a similar problem, I'm just going to use that as a guide http://math.berkeley.edu/~borisp/MA104/MA104solutions3.pdf

Answer 51

Thanks again Ganeshie, Sour, and Solomon

Answer 52

\[f'(x) = 4x\sin(1/x) - 2\cos(1/x) +1\]

Answer 53

consider the limit x->0 and show that the derivative function oscialletes "infinitely" between +ve and -ve values

Answer 54

that proves there exists no interval containing 0 in which the function is increasing

(also it proves there is no interval containing 0 in which the function is decreasong)

Answer 55

ok, it was a technicality I didn't understand. If it oscilates, I assumed that meant it both inc and dec

Answer 56

Ohk.. that pdf looks great!

Answer 57

stick to the definition of "increasing function"

Answer 58

you need to have an "interval" for which the derivative must "maintain" its sign for the entire interval

Answer 59

Yea, it does it the same way you did. You go smarty pants! :)

Answer 60

ohhh maintain... that makes sense now

Answer 61

by proving the derivative is "infinitely oscillating" around 0, we are proving that there is no "interval" around 0 in which the derivative "maintains" its sign

Answer 62

That makes sense now. I did not realize that definition of increasing.