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liizzyliizz

  • 4 years ago

find the equation of the tangent line to the curve f(x)= lnx/e^x at x=1

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  1. anonymous
    • 4 years ago
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    slope m=f'(1) point (1,f(1)) equation: (y-f(1))=m(x-1)

  2. TuringTest
    • 4 years ago
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    ^right, so what have you got for f'(x) ?

  3. liizzyliizz
    • 4 years ago
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    wouldn't f'(x) be (1/x - ln(x)) * e^-x ?

  4. TuringTest
    • 4 years ago
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    yeah, so what is f'(1) ?

  5. liizzyliizz
    • 4 years ago
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    it would be 1

  6. liizzyliizz
    • 4 years ago
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    ?

  7. TuringTest
    • 4 years ago
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    \[f'(x)=e^{-x}(\frac1x-\ln x)\]\[f'(1)=e^{-1}(\frac11-\ln 1)=e^{-1}(1-0)=e^{-1}\]

  8. liizzyliizz
    • 4 years ago
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    err I messed up on the e^-1 :/

  9. TuringTest
    • 4 years ago
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    so that's your slope, m what's f(1) ?

  10. liizzyliizz
    • 4 years ago
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    0

  11. anonymous
    • 4 years ago
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    right, so now you have: slope m=e^-1 and point (1,0) replace in the equation

  12. liizzyliizz
    • 4 years ago
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    ok so it would be y-0=e^-1(x-1) correct?

  13. liizzyliizz
    • 4 years ago
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    then you would just change the form, which is what im doing now to match the choices in my question

  14. TuringTest
    • 4 years ago
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    yeah that's right don't know what for you need it in...

  15. TuringTest
    • 4 years ago
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    what form*

  16. anonymous
    • 4 years ago
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    y=e^-1(x-1) is the equation of the tangent line to the curve

  17. liizzyliizz
    • 4 years ago
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    these are the choices- x-ey-1 =0 x+ey-1=0 x-y-1=0 ex+y-1=0 ex-y-1=0

  18. liizzyliizz
    • 4 years ago
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    i feel like when i took this test i did the work somewhat right, but when it came to switching it out i made a careless mistake so i want to know which one was it. :c

  19. TuringTest
    • 4 years ago
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    multiply both sides by e I meant :/

  20. TuringTest
    • 4 years ago
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    actually something is missing...

  21. TuringTest
    • 4 years ago
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    \[\large y=e^{-1}(x-1)\]\[\large ey=x-1\]\[\large x-ey-1=0\]but that's not a choice...

  22. TuringTest
    • 4 years ago
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    oh, it is the first one I didn't see that one on the first line

  23. anonymous
    • 4 years ago
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    liizzyliizz the first choice is -x-ey-1 =0? or x-ey-1 =0?

  24. liizzyliizz
    • 4 years ago
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    first choice is x-ey-1 =0

  25. anonymous
    • 4 years ago
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    that is the answer

  26. liizzyliizz
    • 4 years ago
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    well this helped thank you, I know what I did wrong. *sigh*

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