Question

quick question about quantifiers and 0 slope

Answer 1

Help interpreting below conditional statement :

\(\large \exists \delta \gt 0 , ~ \forall \epsilon \gt 0 : |x-a| \lt \delta \implies |f(x)-f(a)|\lt \epsilon\)

Answer 2

it says there exists some \(\delta\) such that the deviation of functions value is within \(\epsilon\) for ANY \(\epsilon\). does that mean the slope of tangent line at x=a is 0 ?

Answer 3

There exists \(\delta\) greater than 0 on value, For all \(\epsilon\) greater than 0, such that Mod of (x-a) is less than \(\delta\) and it implies that,, ,,,,,,,,, Oh going so long in writing it.. :P

Answer 4

*in value..

Implying Mod of (f(x) - f(a)) less than \( \epsilon\)..

Answer 5

yeah thats right, that means there exists some \(\delta\) which works for all \(\epsilon\) no matter how small the \(\epsilon\) is ?

Answer 6

Can that statement be a sufficient condition for \(\large f'(a) = 0\) ?

Answer 7

yeah but epsilon will be positive only..

Answer 8

actually I got that conditional statement from continuity definition of a function, so yes \(\large \epsilon \gt 0\) sorry look i didn't specify that..

Answer 9

I am really not good at it, but can you tell in what respect you are saying the slope to be \(0\) ? How are you interpreting it?

Answer 10

http://assets.openstudy.com/updates/attachments/5445e5afe4b051a66734eac8-jasonjohnson86-1413866945600-image11.jpg

Answer 11

part c

Answer 12

the difference between f(x) and f(a) should be within \(\epsilon\) and i can choose the value as close to 0 as i want... so the function will not be changing for sometime...

so im thinking part c in that attachment gives some condition using limits to check \(\large f'(a) = 0\)

Answer 13

http://math.stackexchange.com/questions/983829/checking-slope-0-at-a-point-for-a-function-using-epsilon-delta-defi

Is this helping you in any way?

Answer 14

that was asked by me yesterday but its a different question

Answer 15

That you asked?

Answer 16

the discussion there was more about interpreting the difference between order of quantifiers, we couldn't conclude on the derivative question

Answer 17

Yes, you are right, let ikram post her point of view too.. :)

Answer 18

ikram are you sure you are typing only?? :P

Answer 19

ugh why OS comments dont work :-|
however this is the delta epsilone definition
for a neighbor of x there is a neibor of f(x) if f(x) is continues

Answer 20

Nice drawing, I will learn "Drawing" from you one day.. :P

Answer 21

:P

Answer 22

you can say Slope is 0, because you can take smallest possible value and there we can assume that slope has not been changed yet. :)

Answer 23

But, we should consult experts too.. :P

Answer 24

i am not good with epsilon delta proofs too,
tell me this : we define stationary `point` as a point at which the derivative is 0. but here we're getting a finite interval over which the funciton is not changing its value.. so... does it still refer to the derivative at point \(\large a\) ?

Answer 25

ok i think part C is telling that is a one to one function ( im not sure about slope thingy ) it dint make sense to me

Answer 26

oh how one to one ?

Answer 27

i have a side question

Answer 28

its a dumb question actually : when f'(a) = 0, the function stays constant briefly right ?

Answer 29

epsilon delta is telling u that a function is continues means for any s in domain lim f(x) as x goes to a is exists

Answer 30

yes , so ?

Answer 31

yes it exists and the limit equals f(a)

Answer 32

keep going, il ask my question again afterwards

Answer 33

Oh yes you're right, its just an existence of limit definition.. like the definition for a sequence to converge. I was not looking at it from that view :o

Answer 34

ok so if u wanna understand epsilon delta thingy , start like this :-
\(\forall \epsilon \gt 0 \exists \delta \gt 0 : |x-a| \lt \delta \implies |f(x)-f(a)|\lt \epsilon\)
means when f(x) is continues then for any a in Domain there exist an interval around a (neighbor )