Finding the local extrema of this function

In summary, the function F(x)=x^m * (1-x)^n where m and n is greater or equal to 2 has local extrema at x = 0, 1, and 1/2, and points of inflection at m/(m+n) and n/(m+n). The exact x coordinates of the local extrema and points of inflection will vary depending on the values of m and n. It is important to take into consideration limit behavior when determining these points.
  • #1
jason_r
27
0
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
 
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  • #2
Well how do you find extrema in general?
 
  • #3
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
 
  • #4
You can still differentiate and group things, hopefully you can factor them into a product of a few things?
 
  • #5
jason_r said:
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:

[tex]x^m = x x^{m-1}[/tex]
 
  • #6
ok i got up to here:

m/(n+m) = x

would that be the only possible x coordinate for a local extrema?
 
  • #7
Well let's let [tex] n = m = 2[/tex] you then have

[tex] f(x) = x^2(1-x)^2 = x^4 - 2x^3 + x^2 \Rightarrow f'(x) = 4x^3 - 6x^2 + 2x [/tex]

If we set that equal to 0 we get

[tex] f'(x) = 2x(2x^2 - 3x + 1) = 0 \Rightarrow x = 0, \frac{1}{2}, 1 [/tex]

This gives one of your responses

[tex] \frac{m}{n+m} [/tex]

Since if we are doing my case we would get

[tex] m = n = 2 \Rightarrow \frac{2}{2+2} = \frac{1}{2} [/tex]

Unless I'm misunderstanding you?
 
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  • #8
so i should assume m=n=2? why is that
also would the answer be 1/2 ?
 
  • #9
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.
 
  • #10
ok using your method i got 3+- root 1/ 4

how did you get i(?)root 7?
 
  • #11
I made a mistake the roots are x = 0, 1, 1/2, so your original solution might be partially right
 
  • #12
jason_r said:
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.
 
  • #13
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.
 
  • #14
NoMoreExams said:
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).
 
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  • #15
Are you sure, maple gives it as one, I'm curious why you don't think it is.
 
  • #16
NoMoreExams said:
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.
 

Related to Finding the local extrema of this function

What is the definition of a local extremum?

A local extremum is a point on a curve where the function's value is either the highest or lowest in the immediate vicinity of that point.

What is the difference between a local maximum and a local minimum?

A local maximum is the highest point on a curve while a local minimum is the lowest point. Both are considered local extrema.

How do you find the local extrema of a function using calculus?

To find the local extrema of a function using calculus, take the derivative of the function and set it equal to zero. Solve for the x-value(s) that make the derivative equal to zero. These x-values are the critical points. Then, plug the critical points into the original function to find the corresponding y-values, which are the local extrema.

What is the difference between a relative and absolute extrema?

A relative extrema is a point on a curve where the function's value is either the highest or lowest in a specific region, while an absolute extrema is the highest or lowest point on the entire curve.

Can a function have multiple local extrema?

Yes, a function can have multiple local extrema. These points can be either relative or absolute extrema.

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