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anonymous

  • 5 years ago

anyone know what dividing ln(x)/(x) does exactly to the graph? so the function is y=ln(x)/(x). I'm supposed to explain how "each algebraic piece of the function affects portions of the graph" so the algebraic piece in this case is the division of x.

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  1. anonymous
    • 5 years ago
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    Graph them individually in a calculator, you would see that the ln x dominates and bends the x

  2. anonymous
    • 5 years ago
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    oh okay wait are you still here? can we discuss this more?

  3. anonymous
    • 5 years ago
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    can you be more specific about it "bends the x" and we need to talk specific portions of the graph so when x < 0 and when x>0 what happens i think

  4. anonymous
    • 5 years ago
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    I'll have do look this up, see if i have any info in old notes.

  5. anonymous
    • 5 years ago
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    okay thanks!

  6. anonymous
    • 5 years ago
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    :)

  7. anonymous
    • 5 years ago
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    find anything? :)

  8. anonymous
    • 5 years ago
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    You have to plug in x values it would give you clues. X values less than zero undefined and negative denominator x --> tends to zero. Do same process other direction.

  9. anonymous
    • 5 years ago
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    wait let me see if I understand so when x < 0 all values are undefined, when x > 0 though..

  10. anonymous
    • 5 years ago
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    All ln values, the denominator is defined.

  11. anonymous
    • 5 years ago
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    when x < 0 on my calculator's table it just says ERROR

  12. anonymous
    • 5 years ago
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    Yes, laws of log. Log equal to or greater than zero

  13. anonymous
    • 5 years ago
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    it looks like after a certain point it pulls the graph down and keeps it close to the x axis compared to just plain y=ln(x)

  14. anonymous
    • 5 years ago
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    Effect of smaller ln x values divided by larger x values in denominator.

  15. anonymous
    • 5 years ago
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    right, ok now im going to type my teacher's example she gave us and we can try to mimick she explained for y=e^(x)+x that when x < 0 the graph of the function is closer to the x part of the function. This is because as x gets more and more negative, e^(x) gets smaller and samller and the function gets closer to x. When x>0 the graph is closer to the e^(x) [art f the function because as x approaches inifinity, e^(x) becomes much larger than x so the graph acts like the e^(x) part.

  16. anonymous
    • 5 years ago
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    Best explanation, I can think of, on the bottom half of the graph it tends to zero because ln x can not be less than zero, points downward due to negative numbers in denominator. Top half, very small values of x divided by very large values of x tends to zero along the y axis to positive infinity.

  17. anonymous
    • 5 years ago
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    oh okayyy

  18. anonymous
    • 5 years ago
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    still there?

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