anonymous
  • anonymous
y= 2x/ (1-x) y'' = 4 / (1-x)^3 I need to find the concavity and the point of inflection so I need to use the second derivative but x=1 is undefined, so does it mean there is no concavity nor point of inflection?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
no. depending on the sign of f'' on the left/right of x=1, will mean concave up/down on the left/right of x=1
anonymous
  • anonymous
and you're right, there is no inflection point. but there might be a change in concavity as you go from left of x=1 to the right of x=1.
anonymous
  • anonymous
it is + in the intervals of ( - infinity , 1) and -ve in ( 1, + infinity) so there is point of inflection?

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anonymous
  • anonymous
but then when i plug in x=1, it is undefined
anonymous
  • anonymous
so there is still concavity even there is no point of inflection?
anonymous
  • anonymous
|dw:1332290926294:dw| take this for example. the function is undefined at x=c but on the left of x=c, the function is concave down. but to the right of x=c, the function is concave up.
anonymous
  • anonymous
yes i get it ! tyvm!
anonymous
  • anonymous
good...
anonymous
  • anonymous
btw, I was answering your question based on your second derivative. I think the derivative should be -4/(x-3)^3.
anonymous
  • anonymous
sorry, -4/(x-1)^3
anonymous
  • anonymous
my teacher gave me that second derivative which is 4/ (1-x)^3
anonymous
  • anonymous
hmm.
anonymous
  • anonymous
so isnt it the same if u take away the -ve sign
anonymous
  • anonymous
still the same.
anonymous
  • anonymous
yea yea I have another question about the vertical tangent but then that one is another function, not this one
anonymous
  • anonymous
go ahead
anonymous
  • anonymous
About vertical tangent If a point (e.g x=1) is undefined in the original f(x) function, that means it is a vertical asymptote right so for f'(x), x=1 is the critical number, and it is also undefined the signs are both positive, which indicates a vertical tangent but that point is already a Vertical asymptote, so it cant be a vertical tangent then when will there be a vertical tangent?
anonymous
  • anonymous
the function is defined (it has a y-value) at the vertical tangent at x=c but the function is not defined at x=c. take this for example of y = x^(1/3) |dw:1332292189103:dw| the vertical tangent occurs at x=0 and the function is defined at x=0. if this was a vertical asymptote, the function would get closer and closer the vertical asymptote but not touch it.
anonymous
  • anonymous
in the first line I mean "but the function is not defined at x=c" if x=c is a vertical asymptote.
anonymous
  • anonymous
yea, so a point cannot be a vertical tangent and a vertical asymptote at the same time? but how do i know whether that point is V.A or V.T do i need to keep plugging in points to see
anonymous
  • anonymous
ohhh i get it now!
anonymous
  • anonymous
right. because a vertical tangent is on the graph.
anonymous
  • anonymous
well touches the graph.
anonymous
  • anonymous
so the point will have a value in f(x) but undefined in f'(x) = vertical tangent the point will be undefined in f(x) and f'(x) = V.A is this right?
anonymous
  • anonymous
yep
anonymous
  • anonymous
ok thanks so much!! wish i can give u one more medal :D u helped me a lot!!
anonymous
  • anonymous
np

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