Finding the local extrema of this function

[jason_r]

F(x)=x^m * (1-x)^n where m and n is greater or equal to 2?
find the x coordinates of local extrema?

would my answer be with respect to m and n? or do i plug in a number for these variables

 

[NoMoreExams]

Well how do you find extrema in general?

 

[jason_r]

take the derivative of the function set equal to zero and solve for x?
then if the function increases at a point less than the critical number and decreses after the critlcal point then its a max, and otherway if its min

but this function has the variables m and n and i can't solve for ciritcal numbers

 

[NoMoreExams]

You can still differentiate and group things, hopefully you can factor them into a product of a few things?

 

[Chalnoth]

take the derivative of the function set equal to zero and solve for x?
then if the function increases at a point less than the critical number and decreses after the critlcal point then its a max, and otherway if its min

but this function has the variables m and n and i can't solve for ciritcal numbers

Right, so your answer will be in terms of m and n. Just figure out the derivative in terms of these two unknown numbers, and set it equal to zero.

A little trick that should help:

$$x^m = x x^{m-1}$$

 

[jason_r]

ok i got up to here:

m/(n+m) = x

would that be the only possible x coordinate for a local extrema?

 

[NoMoreExams]

Well let's let $$n = m = 2$$ you then have

$$f(x) = x^2(1-x)^2 = x^4 - 2x^3 + x^2 \Rightarrow f'(x) = 4x^3 - 6x^2 + 2x$$

If we set that equal to 0 we get

$$f'(x) = 2x(2x^2 - 3x + 1) = 0 \Rightarrow x = 0, \frac{1}{2}, 1$$

This gives one of your responses

$$\frac{m}{n+m}$$

Since if we are doing my case we would get

$$m = n = 2 \Rightarrow \frac{2}{2+2} = \frac{1}{2}$$

Unless I'm misunderstanding you?

 

[jason_r]

so i should assume m=n=2? why is that
also would the answer be 1/2 ?

 

[NoMoreExams]

I was picking an example, if your answer is right it should work for ALL (m,n). You said the only restriction was that m, n has to be greater than 2 and so I picked the simplest example. Using basic calculus I got one set of answers, using YOUR answer I got a different one, that shouldn't happen. This is why it's important to show your work instead of just giving an answer you think is right.

 

[jason_r]

ok using your method i got 3+- root 1/ 4

how did you get i(?)root 7?

 

[NoMoreExams]

I made a mistake the roots are x = 0, 1, 1/2, so your original solution might be partially right

 

[Chalnoth]

ok i got up to here:

m/(n+m) = x

would that be the only possible x coordinate for a local extrema?

You definitely made a mistake here.

First, you should have three different roots. So you appear to have taken the one interesting-looking root and just ignored the other two. Make sure to get all of them.

Second, this root you have here looks very similar to one of the three roots, but it isn't quite right. I'm not quite sure where your error is, but it's probably a very, very simple one.

 

[NoMoreExams]

m/(m+n) is local extrema. 0 and 1 are points of inflection I believe. Unless you define your function on a closed subset of R... you need to look at limit behavior too.

 

[Chalnoth]

m/(m+n) is local extrema. 0 and 1 are points of inflection I believe. Unless you define your function on a closed subset of R... you need to look at limit behavior too.

m/(m+n) isn't quite correct, though. And I'm pretty sure that whether or not any of the roots is a point of inflection depends upon m and n, though I could be mistaken there.

Edit: Yes, I went back and checked, and whether or not m or n are points of inflection depends upon m and n. I also verified that m/(m+n) is not an extremum (but it's close).

 

[NoMoreExams]

Are you sure, maple gives it as one, I'm curious why you don't think it is.

 

[Chalnoth]

Are you sure, maple gives it as one, I'm curious why you don't think it is.

Ah, my bad. I wrote down the problem wrong (transposed m and n). Sorry about any confusion.