anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Graph them individually in a calculator, you would see that the ln x dominates and bends the x
anonymous
  • anonymous
oh okay wait are you still here? can we discuss this more?
anonymous
  • anonymous
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

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anonymous
  • anonymous
I'll have do look this up, see if i have any info in old notes.
anonymous
  • anonymous
okay thanks!
anonymous
  • anonymous
:)
anonymous
  • anonymous
find anything? :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
wait let me see if I understand so when x < 0 all values are undefined, when x > 0 though..
anonymous
  • anonymous
All ln values, the denominator is defined.
anonymous
  • anonymous
when x < 0 on my calculator's table it just says ERROR
anonymous
  • anonymous
Yes, laws of log. Log equal to or greater than zero
anonymous
  • anonymous
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)
anonymous
  • anonymous
Effect of smaller ln x values divided by larger x values in denominator.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh okayyy
anonymous
  • anonymous
still there?

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