Candidate Functions f(x) for Domain R, Image (-1,1)

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In summary, some suitable candidate functions for f(x) that satisfy the given conditions are rational functions, sigmoid curves, and a function of the form f(x)=C√[arctan(|x|)], where C is a constant. However, a rational function would not work as it has the same asymptote at both ends, while the other two options may not have a disappearing first derivative. Therefore, taking the cube root of arctan(x) may be the best solution as it has the desired vertical slope at x=0.
  • #1
longrob
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Hi all

What are some candidate functions f(x) that satisfy these conditions:
1. domain of f is R
2. image of f is (-1,1)
2. Smooth and continuous everywhere
3. first derivative undefined at x=0
4. f(x)-->1 as x--> inf
5. f(x)-->-1 as x--> -inf

Thanks
LR
 
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  • #2
Well the horizontal asymptotes are at 1 and - 1, so a suitable rational function should do the trick.
 
  • #3
Thanks, but I've not been able to find one. Any suggestions ?
 
  • #4
check sigmoid curves
 
  • #5
Thanks. I considered arctan already, but since this function goes momentarily vertical zero arctan doesn't work. Same with a Gompertz function and Richards curve (I think). Also, this function appears to be odd, so that would rule out a Gompterz function also. Are there Sigmoid curves that are odd ? I don't know much about them, except in population models, and in those models a disappearing first derivative isn't too desirable I guess.
 
  • #6
Well, you might try with something like:

x>0: [itex]f(x)=C\sqrt{arctan(|x|)}[/itex]
x<0: [itex]f(x)=-C\sqrt{arctan(|x|)}, C=\sqrt{\frac{2}{\pi}}[/itex]
 
  • #7
snipez90 said:
Well the horizontal asymptotes are at 1 and - 1, so a suitable rational function should do the trick.

No, a rational function has the same asymptote at both ends.

So: in rejecting arctan, you say that you WANT it to be vertical at zero? (arctan has slope 1 at zero). Then take the cube root: [tex](arctan(x))^{1/3}[/tex]
 

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  • #8
g_edgar said:
So: in rejecting arctan, you say that you WANT it to be vertical at zero? (arctan has slope 1 at zero). Then take the cube root: [tex](arctan(x))^{1/3}[/tex]

Perfect. Thank you.
 

1. What is a candidate function f(x) for a domain of R and an image of (-1,1)?

A candidate function f(x) is a mathematical expression or rule that takes in any real number as input (known as the domain) and produces an output between -1 and 1 (known as the image). In other words, it is a function that maps the real numbers to a specific range of values.

2. How do you determine if a function is a candidate function for a given domain and image?

To determine if a function is a candidate function for a given domain and image, you need to plug in different values for x in the function and check if the resulting outputs fall within the specified image range of (-1,1). If the outputs are all within this range, then the function is a candidate function for the given domain and image.

3. Can a candidate function have a domain and image that are both equal to R?

Yes, a candidate function can have a domain and image that are both equal to R, as long as the function's outputs still fall within the specified range of (-1,1).

4. What is the importance of finding candidate functions for a given domain and image?

Finding candidate functions for a given domain and image is important in many areas of mathematics and science. It allows us to model real-world situations, make predictions, and solve problems by using mathematical expressions. It also helps us understand the relationships between different variables and how they affect each other.

5. Are there any restrictions on the types of functions that can be candidate functions for a domain of R and an image of (-1,1)?

There are no specific restrictions on the types of functions that can be candidate functions for a domain of R and an image of (-1,1). However, the function must be defined for all real numbers and its outputs must fall within the specified range of (-1,1) for it to be considered a candidate function for the given domain and image.

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