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Flofloflo
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Joined: 07 Nov 2007
Posts: 120
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 Posted: 11 Apr 2008 08:44:00 am Post subject: |
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Hello,
Today, at school, I had a nearly impossible to make math test... at least, nobody in my class seemed to understand anything of it, and we are on the highest level there is in Netherlands FYI:P
I am not exactly sure how to put this in English, but we had to find the Tangent of the bending point of a formula; by using the second Derivative.... Y'all should know what I mean now, right?
Oh yeah, and only use algebra
The formula was f(x)=(lnx)^2+2LnX-2
My solution:
f'(x)=(2LnX)/X+2/X
f''(x)=(2-2ln(x))/x^2-(2/X^2) = (-2ln(x)/(x^2)
Now, if I am right, I should make f''(x) = 0.
To accomplish that: x=1.
THat means the bending point is (-2,1), by putting it in the F(X).
To get the tangent, you need the function of a line : Y=aX+b
To get a you put the x of the point in the f'(x); that returns 2.
so the formula is Y=2X+B. I goes trough point (-2,1) As I earlier said; so -2=2+B. B=-4.
So I think the formula should be Y=2X-4. Am I wright??
The next question was this, about the same formula:
There are two lines that go trough point 0,0 that are tangents of the original formula. Where do these two lines tough the formula?
I have absolutely no idea... Anyone?>?>
I hope I descirbed the problem decently.... Maybe not, I am not ENglish after all-_-
Last edited by Guest on 11 Apr 2008 08:44:43 am; edited 1 time in total |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 11 Apr 2008 01:40:00 pm Post subject: |
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Your description is all right, except the correct terminology is "point of inflection" in English (the same thing, but in fancy language).
As for the next question: you know a line through the origin has a formula y=kx. If it's tangent at the point of intersection, then k=f'(x) so the point of intersection has y=xf'(x).
Now set f(x) equal to xf'(x) and substitute for the values of f(x) and f'(x):
(lnx)^2+2ln(x)-2 = x(2ln(x)+2)/x
This should give you two solutions for x. The lines tangent at those points should pass through the origin. |
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Flofloflo
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Joined: 07 Nov 2007
Posts: 120
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| Posted: 12 Apr 2008 10:11:09 am Post subject: |
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Hmm... I understand actually... But I would have never thought of that... Atlough it's actually not that difficult........
Oh, and my solution to the first question was correct, right? |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 12 Apr 2008 01:23:55 pm Post subject: |
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| Yeah, the tangent line at (1,-2) is y=2x-4 |
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Flofloflo
Member
Joined: 07 Nov 2007
Posts: 120
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| Posted: 13 Apr 2008 05:20:39 am Post subject: |
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DarkerLine wrote: Your description is all right, except the correct terminology is "point of inflection" in English (the same thing, but in fancy language).
As for the next question: you know a line through the origin has a formula y=kx. If it's tangent at the point of intersection, then k=f'(x) so the point of intersection has y=xf'(x).
Now set f(x) equal to xf'(x) and substitute for the values of f(x) and f'(x):
(lnx)^2+2ln(x)-2 = x(2ln(x)+2)/x
This should give you two solutions for x. The lines tangent at those points should pass through the origin.
[post="122391"]<{POST_SNAPBACK}>[/post]
I might be wrong, but you seem to be replacing the k in y=kx by x when you say x(2ln(x)+2)/x.
Why's that? |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 13 Apr 2008 10:06:56 am Post subject: |
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| I'm not. I'm replacing the k by (2ln(x)+2)/x, which is f'(x). It's just that I put the x on the other side for some reason, sorry if that was confusing. |
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