anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
prove it if it is true or if it is false give a short proof for saying so.
amistre64
  • amistre64
is there a typo in that question? f'(1/2) exists but does not exist?
anonymous
  • anonymous
i'm also confused.

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amistre64
  • amistre64
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
anonymous
  • anonymous
sorry, I got the mistake plz replace 'not exixts' with ' not an element of'
amistre64
  • amistre64
1/0 or 0/0 or inf/inf or some other setup
amistre64
  • amistre64
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...
anonymous
  • anonymous
ok, it seems so.
amistre64
  • amistre64
it sounds like a case for a rational function..... but i cant put my finger on it yet
amistre64
  • amistre64
or can we define "peicewise" functions as f(x)?
anonymous
  • anonymous
no
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
(x+3)(x-(1/2)) ------------ (x-2)(x-(1/2)) This function has a "hole" at x=1/2. I wonder if it counts

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