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nshiba Group Title

e^x=-x+1, how do you solve this? @MIT 18.01 Single …

  • 2 years ago
  • 2 years ago

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  1. JamesJ Group Title
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    In this case x = 0 is the solution.

    • 2 years ago
  2. JamesJ Group Title
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    Normally with an expression e^(ax) = polynomial in x, numerical methods are needed. But with yours we can see a solution by inspection.

    • 2 years ago
  3. arijit.mech Group Title
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    friend it is easy:just draw the graphs of y=e^x & y=-x+1....>the intersection point is the solution....this is a transcendental equation |dw:1320213287494:dw|

    • 2 years ago
  4. rosswante Group Title
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    You may see http://www.wolframalpha.com/input/?i=solve%7Be%5Ex%3D-x%2B1%7D

    • 2 years ago
  5. arijit.mech Group Title
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    Thanks....i jst give u a hint

    • 2 years ago
  6. verma.milan Group Title
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    mathematically take natural log on both sides e u will get point x where the solution exists

    • 2 years ago
  7. verma.milan Group Title
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    |dw:1321869467155:dw|

    • 2 years ago
  8. alireza_masoomi Group Title
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    dif ratio x then Ln that result x=e^-1

    • 2 years ago
  9. alireza_masoomi Group Title
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    hi! e^x=-x+1when dif it then e^x=-1 after Ln xLne=Ln-1 that result x=e^-1 ~ .36

    • 2 years ago
  10. alireza_masoomi Group Title
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    |dw:1322221990014:dw||dw:1322222039311:dw|

    • 2 years ago
  11. JamesJ Group Title
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    The equation \[ e^x = -x + 1 \] has one and only solution, \( x = 1. \ \ \ x = e^{-1} \) is not a solution, something that is easily verifiable by substitution. (The uniqueness of the solution \( x = 1 \) can be proven by examining the function \( f(x) = e^x + x - 1. \ \ \ x = 1 \) is clearly a zero of f, but as \( f'(x) = e^x + 1 > 0 \) for all \(x\), it follows that it is also the only zero.)

    • 2 years ago
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