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

[312213]

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.

 

[jackmell]

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.

 

[312213]

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.

 

[Inferior89]

You can't get it in an exact form.

http://www.wolframalpha.com/input/?i=inflexion+Si(x),+-10<x<10

http://www.wolframalpha.com/input/?i=tan(x)+=+x

You see they have the same numerical solutions so your work is correct.

 

[312213]

So then an acceptable answer would be (tan(x),Si(tan(x))?

 

[Inferior89]

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

 

[312213]

I see and understand now. Thanks for the help.

 

[Bohrok]

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.

 

[Inferior89]

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.

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