Find Function Like in Picture - No Trig Needed

[TalonStriker]

Can anyone give me a function that (roughly) looks like the top or bottom one in the picture attached? And yes the limits is constant as x-> +/- infinity. I don't want a trig function like arctan since trig isn't really all that applicable for what I'm doing.

I'm posting here since I definitely saw functions like this in my deff eq class, but have unfortunately forgotten its formula.

Thanks.

 

[arildno]

Well, the hyperbolic tangent function can achieve such behaviour easily, with addition of a constant, sign choice and scaling.
Similarly with the arctan function

 

[TalonStriker]

Well, the hyperbolic tangent function can achieve such behaviour easily, with addition of a constant, sign choice and scaling.
Similarly with the arctan function

you missed the part where I said that I didn't want trig functions.

 

[Hootenanny]

you missed the part where I said that I didn't want trig functions.

tanh isn't a trigonometric function.

 

[sutupidmath]

what about

f(x)=- x^{\frac{1}{3}},if,x\geq 0;(-x)^{\frac{1}{3}}, if, x<0 this is a piecewise defined funct.

 

[statdad]

Look at the class of logistic functions.

 

[sutupidmath]

what about

f(x)=- x^{\frac{1}{3}},if,x\geq 0;(-x)^{\frac{1}{3}}, if, x<0 this is a piecewise defined funct.

Mine doesn't really work, since i didn't see the restriction that it has to have a horizontal asymptote.

 

[Integral]

The http://en.wikipedia.org/wiki/Error_function" come close to what you want.

 

[arildno]

I don't want a trig function like arctan since trig isn't really all that applicable for what I'm doing.

you missed the part where I said that I didn't want trig functions.

What nonsense.
Do you think you can't "apply" an arctan function just because you are not dealing with a problem in trigonometry?

 

[Count Iblis]

And yes the limits is constant as x-> +/- infinity

Have you got accurate numerical values or just the shape of the graph? If you got numerical data you could take a shot at exactly guessing the function.

 

[St. Aegis]

just use integral's curves lawl