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lgbasallote Group Title

find f \[f''(x) = x^{-2}, \quad x > 0,\quad f(1) = 0, \quad f(2) = 0\]

  • one year ago
  • one year ago

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  1. myininaya Group Title
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    To find f you need to first find f' the find f' given f'' integrate both sides

    • one year ago
  2. lgbasallote Group Title
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    hmm.. \[f'(x) = \frac 1x + c\] ??

    • one year ago
  3. myininaya Group Title
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    almost

    • one year ago
  4. myininaya Group Title
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    you are missing a certain constant multiple

    • one year ago
  5. lgbasallote Group Title
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    ahh \[f'(x) = -\frac 1x + c\]

    • one year ago
  6. myininaya Group Title
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    yes :) now to find the constant hmmm....you are missing a certain initial condition to do that .... your question doesn't make sense .... you need one of those to be f'(something)=another something

    • one year ago
  7. lgbasallote Group Title
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    or maybe i should do it \[f'(x) = -\frac 1x + c_1\]

    • one year ago
  8. lgbasallote Group Title
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    i might have typoed

    • one year ago
  9. Zarkon Group Title
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    integrate again

    • one year ago
  10. lgbasallote Group Title
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    ah yes i did. it's f'(2) not just f(2)

    • one year ago
  11. Zarkon Group Title
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    get a system of 2 eau and 2 unknowns

    • one year ago
  12. lgbasallote Group Title
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    hmm \[f(x) = -\ln x + c_1 x + c_2\] yes?

    • one year ago
  13. myininaya Group Title
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    ok great. you can find that constant by using f(1)=0 and then do what zarkon says to find f

    • one year ago
  14. Zarkon Group Title
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    you can do the problem with two given values of f

    • one year ago
  15. myininaya Group Title
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    or you can wait to find the first constant whatever

    • one year ago
  16. lgbasallote Group Title
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    i suppose x > 0 is just there to note that -ln x exists?

    • one year ago
  17. myininaya Group Title
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    oh wait.... i guess you can do it with f(something1)=another something1 and f(something2)=another something2 oops

    • one year ago
  18. lgbasallote Group Title
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    f'(2) = 0 so f'(2) = -1/2 + c_1 = 0 does this mean c_1 is 1/2?

    • one year ago
  19. myininaya Group Title
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    yes adding 1/2 to both sides solves that equation for c_1

    • one year ago
  20. lgbasallote Group Title
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    then f(1) = -ln (1) + 1/2 x + c_2 = 0 so c_2 is -1/2?

    • one year ago
  21. myininaya Group Title
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    x is 1 so you have -ln(1)+1/2(1)+c_2=0 and yes

    • one year ago
  22. lgbasallote Group Title
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    oh. yeah...forgot to sub 1 into x the second time

    • one year ago
  23. lgbasallote Group Title
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    so \[f(x) = -\ln x + \frac 12 x - \frac 12\] ??

    • one year ago
  24. myininaya Group Title
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    tep

    • one year ago
  25. lgbasallote Group Title
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    nice. thanks

    • one year ago
  26. lgbasallote Group Title
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    i just noticed this was my 1000th question

    • one year ago
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