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anonymous

  • 5 years ago

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.

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  1. anonymous
    • 5 years ago
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    prove it if it is true or if it is false give a short proof for saying so.

  2. amistre64
    • 5 years ago
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    is there a typo in that question? f'(1/2) exists but does not exist?

  3. anonymous
    • 5 years ago
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    i'm also confused.

  4. amistre64
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    sorry, I got the mistake plz replace 'not exixts' with ' not an element of'

  6. amistre64
    • 5 years ago
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    1/0 or 0/0 or inf/inf or some other setup

  7. amistre64
    • 5 years ago
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    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...

  8. anonymous
    • 5 years ago
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    ok, it seems so.

  9. amistre64
    • 5 years ago
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    it sounds like a case for a rational function..... but i cant put my finger on it yet

  10. amistre64
    • 5 years ago
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    or can we define "peicewise" functions as f(x)?

  11. anonymous
    • 5 years ago
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    no

  12. amistre64
    • 5 years ago
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    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 :)

  13. amistre64
    • 5 years ago
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    (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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