Question

f(x) = | x | is contiguous function ?
i will provide some images so that u can help me easily...

Answer 1

it is continuous yes, not differentiable at x = 0

Answer 2

you perl u r absolutely right.... plz can u explain how it is contiguous.... fun

Answer 3

Sorry no. :l

Answer 4

lgba.... thanks for coming in here.. plz help me

Answer 5

huh.if saifoo doesnt know it...i guess i dont stand a chance :/ sorry

Answer 6

omg... you mean saifoo knows more than u know .. ?

Answer 7

saifoo... i didn't got u :) :)

Answer 8

im saying that saifoo knows more calculus than me...but in all other aspects of math...he's just a faster typer :P

Answer 9

@jatinbansalhot , Typo!
i was saying;
lol.. what should i say..?

Answer 10

lgb.... then saifoo could have helped me but he didn't maybe he is short by time :) :)

Answer 11

is |x| continuous function at x = 0 :) :)

Answer 12

i also hate the graphing part of calculus :P i only liked the differentiating and integrating

Answer 13

Haha. @lg

@J
But Jatin, sorry i really dont have any idea about this.
But i will try to get you help.

Answer 14

@apoorvk HELP!

Answer 15

Saifoo.. thanks ....
lagb .. then u can help me ... tell me plz |x| is derivative at x = 0

Answer 16

apoorvkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk :) :) eventually u here :) :)

Answer 17

|x| is approximately 143.6% a continuous function.
(one shortcut of determining whether a functionis continuous or not, is see if while plotting the entire, you have to 'lift' you pen or not.)

Answer 18

Differentiability is another issue altogether. There is a 'sharp point' at x=0, so |x| isn't differentiable at x=0.

Answer 19

apoorvk...... at x = 0 how we can prove that it is contiguous ?

Answer 20

continuous everywhere and differentiable everywhere except at x=0

Answer 21

graphically we can understand but my tution teacher told me to prove it.

Answer 22

i need proof... how plz

Answer 23

simple, |f(x) - 0| < epsilon implies |x - 0| < delta implies limit = 0 at x=0
f(0) = 0 => so continuous

Answer 24

Okay, at x=0, the function equals zero right?
At its LHL, the function is almost zero as well.
At RHL too, the function is almost 0.

so
lim f(x) =f(0)
x->0


So, mathematically as well, its proved continuous.

Answer 25

not differentiable because differentiable implies the tangency at that point,
now think of how many tangent you can possibly draw at that point??

Answer 26

hahaahahahahaahah.. @experimentX i'm laughing alot.. so simple for u na.. but for me so so so difficult to understand that proof.. :) :)

Answer 27

sorry, differentiable implies the left tangent must be equal to righ tangent => only one tangent at that point.

but there are many (infinite tangent) at that point.

Answer 28

it's usual way ... you find it hard because you barely find function that is continuous but not differentiable.

Answer 29

A function is continuous at a point, if its RHL = LHL = f(at that point)

Answer 30

ohhh.. there are many at that point.. now i got a little.. i thought we can draw a tangent at x = 0 then it ,must be differentiate at that point..

Answer 31

oh, there's one another example x sin(1/x) <--- continuous but not differentiable at x=0

Answer 32

exactly jatin, now you're getting he hang of it!

Answer 33

Hm .. how about the traditional light hand limit = right hand limit = f(a) test?

Answer 34

@jatin ,,, you must really see the definition of differentiation

lim h->0+ f(x+h) - f(x)/h = lim h->0- f(x+h) - f(x)/h (condition)
but,
lim h->0+ f(x+h) - f(x)/h != lim h->0- f(x+h) - f(x)/h (in our case)

Answer 35

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/MIT18_01SCF10_ex02sol.pdf

Answer 36

i understood lim h->0 ( f(x+h) - f(x) ) /h but can't understand
lim h->0 ( f(x+h) - f(x) ) /h

Answer 37

I suggest you to go through this: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/

Prof. Jerison is just amazing!

Answer 38

i understood lim h->0 ( f(x+h) - f(x) ) /h but can't understand lim h->0 (- f(x+h) - f(x) ) /h

Answer 39

i understood lim h->0+ ( f(x+h) - f(x) ) /h h is positive,
f(x+h) is also positive, so both numerator is also positive

lim h->0- ( f(x+h) - f(x) ) /h numerator is positive, h is negative, so the whole value is negative.

Answer 40

yeah ... one is right handed, h is positive,
the other is left handed, so h is negative in this case.

Answer 41

ohhhh.... ...

Answer 42

LOL ... i even managed to copy "i understood "

Answer 43

hahahahahahaha...........@experimentX :) ..... could i start "single variable calculus"..
Are "single variable calculus" and "Calculus 1" same... ?

Answer 44

yeah ... i hope you understand.

Answer 45

thanks .. :) @experimentX @apoorvk @FoolForMath and all others for really help me

Answer 46

all right medal for everyone!!