# Limit The Range Of A FunctionWatch

#1
Suppose we have a function y = f(x) which has a domain such that x >= 0 and xER (x is an element of real numbers). Is there any way I can make the range be 0 < y < 1 and that the bigger x is the bigger y is?
0
quote
11 years ago
#2
yes. of the top of my head, y = (1+tanhx)/2 would work.

all you need to do is think of a one-to-one function with finite limits, and transform it to suit..
0
quote
11 years ago
#3
Or (2/pi)arctan x.

Edit: oh, x in the reals. Erm, ok... (pi/2 + arctan x)/pi. Or something.
0
quote
#4
chewwy, your one reaches 1 too quickly.

generalebriety, your one reaches 1 too quickly as well but if I divide by let's say 3pi then the new problem is that it increases too slowly.

If it helps 0 <= x < infinity.
0
quote
11 years ago
#5
(Original post by DeathAwaitsU)
chewwy, your one reaches 1 too quickly.

generalebriety, your one reaches 1 too quickly as well but if I divide by let's say 3pi then the new problem is that it increases too slowly.

If it helps 0 <= x < infinity.
What are you on about, "too quickly"?
0
quote
11 years ago
#6
i don't understand... too quickly?

y = (x + (1+x^2)^0.5 - 1)/2x is a fine answer too.
0
quote
#7
What I mean is, if I let x = 100, it's too close to 1. I need it to sort of increase a bit slower. I know that there an infinite number of numbers so what I'm saying is stupid out of context but is there a way I could multiply a part of your function by a constant to make it work in context? The constant would sort of control the speed of increase.
0
quote
11 years ago
#8
y = (1+tanh(x/10000))/2
0
quote
#9
(Original post by SsEe)
y = (1+tanh(x/10000))/2
I like that but, I know I've now changed the conditions from what I initially posted, at x = 0 it's outputting 0.5 when I'd want it to output 0.
0
quote
11 years ago
#10

You can then choose k to be where the function hits 0.5 as an easy to compute measure of how "fast" the function approaches 1. It's probably a faster function to calculate than some of the other suggestions if that matters.
0
quote
#11
Thanks, that should work I think.
0
quote
11 years ago
#12
(Original post by DeathAwaitsU)
I like that but, I know I've now changed the conditions from what I initially posted, at x = 0 it's outputting 0.5 when I'd want it to output 0.
You said the domain of this function was R. Of course it's outputting 0.5, we all expected that you wanted it to tend to 0 as x tended to -infinity. Otherwise, my original answer of (2/pi)arctan x works fine. If you want to scale it, (2/pi)arctan(x/10) or whatever.
0
quote
#13
Yeah I know, generale, I forgot I x>=0 when I first posted this. To the person who neg repped me for changing the conditions, I think that was a bit extreme....
0
quote
11 years ago
#14
(Original post by DeathAwaitsU)
Yeah I know, generale, I forgot I x>=0 when I first posted this. To the person who neg repped me for changing the conditions, I think that was a bit extreme but ok....
*shrugs* People will neg rep for all sorts. To be fair to them though, it's very annoying to be asked a question sometimes and then get told you've answered the wrong one.
0
quote
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18
• University of East Anglia
Fri, 4 Jan '19
• Bournemouth University
Wed, 9 Jan '19

### Poll

Join the discussion

Yes (157)
27.3%
No (418)
72.7%