Math Is Fun Forum

Carisma

Member

Registered: 2010-01-08

Posts: 10

Points of Inflection..

I'm just a bit confused with this question, I've found the second derivative, but it looks more like a maximum to me than a point of inflection:

Consider the curve with equation f(x) = e^(-2x^2) for x<0. Find the co-ordinates of the point of inflection and justify that it is a point of inflection.

Thank you for any help in advance!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Points of Inflection..

Hi Carisma;

You haven't told me what points you got for the inflection points.

This one looks fairly easy if you follow the definitions.

http://en.wikipedia.org/wiki/Inflection_point

Here is a little more, for f(c) to be an inflection point then:

I see 2 points that fit the above definition. What did you get?

All that looks pretty tough. I do this:

1) Find the second derivative.

2) Set it to 0, and solve.

3) Now you have possible inflections points.

4) check the points for concavity.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Carisma

Member

Registered: 2010-01-08

Posts: 10

Re: Points of Inflection..

Hi,

I found the second differential, solved for 0 then found the y- coordinate and I got the 2 points of inflection to be (0.5,0.606) and (-0.5, 0.606)

Is this correct?

And thank you for helping

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Points of Inflection..

Hi;

Yes, those are the possible points of inflection. There is more to do.

The x's are correct I don't know where you are getting those y values from. But it is not important.

Check those x's for concavity changes in f''(x).

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: Points of Inflection..

Hi Carisma,

At a maximum the gradient changes from + to - as you pass through the point, and from - to + for a minimum.

At a point of inflexion the gradient doesn't change sign as you traverse the point.

So once you have the differentiated function, checking the gradient just left of the potential point and just right will give you definitive evidence. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Carisma

Member

Registered: 2010-01-08

Posts: 10

Re: Points of Inflection..

Hi;

Yes, those are the possible points of inflection. There is more to do.

The x's are correct I don't know where you are getting those y values from. But it is not important.

Check those x's for concavity changes in f''(x).

I found the y value by substituting the x values into f(x)...am I doing this wrong?

I have checked the concavity on both sides of the x values and it is positive on both sides when x = 0.5 and when x=-0.5

Carisma

Member

Registered: 2010-01-08

Posts: 10

Re: Points of Inflection..

Hi Carisma,

At a maximum the gradient changes from + to - as you pass through the point, and from - to + for a minimum.

At a point of inflexion the gradient doesn't change sign as you traverse the point.

So once you have the differentiated function, checking the gradient just left of the potential point and just right will give you definitive evidence. 

Bob

Thanks Bob! This cleared up my confusion with the differences in concavity around maximum/minimum points and points of inflection

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Points of Inflection..

Hi Carisma;

Yes, that is correct. Did you test both points for concavity changes in f''(x)? It is really easy.

You must perturb x = 1 /2 and x = - 1/2. In other words try two numbers on either side of them.
For instance for 1/2 try 1/4 and 3/4. That is x + 1/4 and x - 1/4. What do you get? Do the same for
x = - 1 / 2. Try for instance -1 / 4 and - 3 / 4. If f''(1/4) and f''(3/4) have different signs
then you have a concavity change.

Take a look at how he handles these. There are other ways but this is simple.

http://www.clas.ucsb.edu/staff/lee/infl … points.htm

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

sameer mishra

Member

Registered: 2010-08-27

Posts: 64

Re: Points of Inflection..

hi carisma
point of inflection is that point where a function changes its nature i.e(it changes its sign from + to - & vice versa)
or mathmatically you can say doubble derivative zero or does not exist at that point

2nd derivative does not exist means in the graph of frist derivative there will be sudden change at that point
in your Q. f(x)= e^(-2x^2)
f"(x)=0 at x=1/2,-1/2
according to condition x=-1/2 is point of inflection