Question

if a function f, defined on [0,1] for which f'{1/2} exists but f'{1/2} not exists [0,1], then f is discontinuous at x=1/2. prove.

Answer 1

prove it if it is true or if it is false give a short proof for saying so.

Answer 2

is there a typo in that question? f'(1/2) exists but does not exist?

Answer 3

i'm also confused.

Answer 4

if it means: f(1/2) exists but its derivative is "indeterminate" meaning that at f'(1/2) there is a vertical line 1/0 ....... can f(x) be continuous? is what I get out of it

Answer 5

sorry, I got the mistake plz replace 'not exixts' with ' not an element of'

Answer 6

1/0 or 0/0 or inf/inf or some other setup

Answer 7

lol .... doesnt help me out any :) the interval itself defines the "domain" and says nothing about the "range" of a function. but maybe im looking at it wrong...

Answer 8

ok, it seems so.

Answer 9

it sounds like a case for a rational function..... but i cant put my finger on it yet

Answer 10

or can we define "peicewise" functions as f(x)?

Answer 11

no

Answer 12

how does the f(x) = 1/x play out in this scenario? decause its derivative is ln(x)...but that doesnt really help me out much.... mainly because i dont really understand the question yet :)

Answer 13

(x+3)(x-(1/2))
------------
(x-2)(x-(1/2))

This function has a "hole" at x=1/2. I wonder if it counts