Thread: Find points of inflection

The curve y= 2x/(1+x^2) has

1. exactly three points of inflection seperated by a point of maximum and a minimum.

2. exactly two points of inflection with a point of maximum lying between them.

3. exactly two points of inflection with a point of minimum lying between them.

4. exactly three points of inflection seperated by two points of maximum.



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My working: I derived the above expression and found the derivative to be 0 at x=1,-1 . So i really dont know how can there be three points of inflection at all.

Hi

Im in kinda of a hurry, but I also just got one inflection point, x=0.
I might have missed something..

Hello,

The second derivative turns out to be :


At an inflection point, the second derivative changes its sign.
If you make a simple table of signs for , you will see that there are indeed 3 inflection points.

The points where the first derivative is 0 are called "critical points". They're not the same as inflection points !
However, finding the first derivative and the critical points will help you answer the parts dealing with maximum/minimum.

Hi Moo.

Is it possible for you to point out the three inflection points in the above graph? Inflection points as I understand are points where the curve changes from concave to convex or vice-versa . Am i correct in this?

Originally Posted by champrock

Hi Moo.

Is it possible for you to point out the three inflection points in the above graph? Inflection points as I understand are points where the curve changes from concave to convex or vice-versa . Am i correct in this?

Yes, you are correct in this. And what I said is an equivalent definition.

Here is the graph.
The inflection points are approximatively where the red circles are.
The -sqrt(3) and sqrt(3) inflection points are not easy to see because the change concave-convex is not very clear (though I found it easy to find it for -sqrt(3))

Did you try that signs table for x(x-sqrt(3))(x+sqrt(3)) ? Actually, a point where a function changes sign is a point where the function is 0. That's logic ^^