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anonymous

  • one year ago

Logistic parent function: 1/1+e^-x. What is the graph and function if 1/+e^-x is shifted left 1 unit? What is the graph and function if flipped over the x-axis?

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  1. anonymous
    • one year ago
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    \[f(x) = \frac{ 1 }{ 1+e^{-x} }\]

  2. SolomonZelman
    • one year ago
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    when you shift a function C units right, you subtract C from every x. when you shift C units left, you add C to every x. --------------------------------------------- To reflect the function across the x-axis, mutliply the [entire] function times 1.

  3. anonymous
    • one year ago
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    So would the function for shifting left one unit horizontally be: \[f(x) = 1/(1+e^x) + 1\]

  4. SolomonZelman
    • one year ago
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    that would be shifted 1 unit up.

  5. anonymous
    • one year ago
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    The one being under the fraction as well.

  6. anonymous
    • one year ago
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    Sorry!

  7. SolomonZelman
    • one year ago
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    you add/subtract the C not from/to the function. you are subtracting/adding to/from the x.

  8. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(x)=\frac{1 }{1+e^{-x}} }\) to shift to the left by 1 unit: \(\large\color{black}{ \displaystyle f(x)=\frac{1 }{1+e^{-(x+1)}} }\)

  9. SolomonZelman
    • one year ago
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    you are adding C to the x (jut as I did). And in this case, the "C" is 1.

  10. anonymous
    • one year ago
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    I see! How would the over x axis function look?

  11. SolomonZelman
    • one year ago
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    you just multiply the entire function times -1, to reflcet over the x-axis.

  12. SolomonZelman
    • one year ago
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    to reflect*

  13. anonymous
    • one year ago
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    So: \[f(x) = -1(\frac{ 1 }{ 1+e^-x }\]

  14. anonymous
    • one year ago
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    Forgot the end parenthesis.

  15. SolomonZelman
    • one year ago
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    Yes, and that would be just same as \(\displaystyle \large \frac{-1}{1+e^{-\left(x+1\right)}}\)

  16. anonymous
    • one year ago
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    Wow! Thank you, I understand now.

  17. SolomonZelman
    • one year ago
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    Anytime!

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