Finding the point of inflection of the integral of (sin(x))/x

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Homework Statement


Find the coordinates of the first inflection point to the right of the origin.


Homework Equations





The Attempt at a Solution


I know that the inflection point would equal zero for the second derivative of a function and that the second derivative of this function is (xcosx-sinx)/x².
(xcosx-sinx)/x² = 0
(xcosx-sinx) = 0
xcosx = sinx
x = tanx

I'm not sure what to do after this or if the steps were taken incorrectly.
 
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312213 said:

Homework Statement


Find the coordinates of the first inflection point to the right of the origin.


Homework Equations





The Attempt at a Solution


I know that the inflection point would equal zero for the second derivative of a function and that the second derivative of this function is (xcosx-sinx)/x².
(xcosx-sinx)/x² = 0
(xcosx-sinx) = 0
xcosx = sinx
x = tanx

I'm not sure what to do after this or if the steps were taken incorrectly.

I don't see what's wrong with that unless you don't know how to then find them numerically.
 
I know that I can find the approximate number but I don't know where to go next to find the exact value of x for the first inflection point.

I don't know what steps to take next to isolate x in one side and to have one x.
 
So then an acceptable answer would be (tan(x),Si(tan(x))?
 
The points of inflexions are when \tan x = x as you said. This equation have an infinite number of solutions.

I don't really understand what you mean with ( \tan(x), Si(\tan(x) ).
One point of inflexion would be for example ( x_1 , Si(x_1)) where x_1 is defined to be the first positive solution to the equation \tan x = x
 
I see and understand now. Thanks for the help.
 
312213 said:
I know that the inflection point would equal zero for the second derivative of a function and that the second derivative of this function is (xcosx-sinx)/x².
(xcosx-sinx)/x² = 0
(xcosx-sinx) = 0
xcosx = sinx
x = tanx

I'm not sure what to do after this or if the steps were taken incorrectly.

[EDIT] Thanks Inferior89

[STRIKE]Actually that's just the first derivative, not the second.
[/STRIKE]
If you want to [STRIKE]consider the values that give the extreme points of sinx/x where you would actually[/STRIKE] solve x = tanx, you might want to take a look here on the second page.
http://press.princeton.edu/books/maor/chapter_10.pdf
Very interesting. :smile:
 
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Bohrok said:
Actually that's just the first derivative, not the second.

If you want to consider the values that give the extreme points of sinx/x where you would actually solve x = tanx, you might want to take a look here on the second page.
http://press.princeton.edu/books/maor/chapter_10.pdf
Very interesting. :smile:

He is finding the second derivative of Si(x) not sin(x)/x.
 
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