Function with f(0)=0,f'(0)=+inf,f(inf)=<1

• sargondjani1
In summary, the function f(x) gives the chance of something happening and that's the reason for f(inf)=<1. The function x^(1/3) - (x-1)^(1/3) + 1 works, but it's not monotonically increasing and it even becomes negative. However, Jacquelin's solution does work perfectly.
sargondjani1
f(x) should give the chance of something happening, that's the reason for f(inf)=<1.

i used f(x)=1-exp(-a*x) until now, which is ok but f'(0) is not +inf. i would prefer a function with f(0)=+inf.

i want the function to be (monotonically) increasing at a decreasing rate (f' decreasing monotonically from +inf to 0)

does a function like this exist?

Play around with xlnx.

When you said $f( \infty )\leq 1$ are you sure that's what you mean? So you could have $f(\infty)=-\infty$?

i don't see how i can get what i want from x ln(x)... can you eloborate? i mean x ln (x) goes to infinity

i meant i want f(inf)=1 (monotonically increasing from 0 to inf.), so f'(inf)=0.

sargondjani1 said:
f(x) should give the chance of something happening, that's the reason for f(inf)=<1.

i used f(x)=1-exp(-a*x) until now, which is ok but f'(0) is not +inf. i would prefer a function with f(0)=+inf.

i want the function to be (monotonically) increasing at a decreasing rate (f' decreasing monotonically from +inf to 0)

does a function like this exist?

Look up the heaviside and dirac delta functions and use these to create your function.

i was shocked at the functions you came up with... then is saw my typo:

i would prefer f'(0)=inf (and not f(0)=inf, since i want f(0)=0)

When I said play around with x lnx, I was thinking of something like (x lnx)/(1 + x2)

chiro said:
Look up the heaviside and dirac delta functions and use these to create your function.

Uuuh, these are distributions. I doubt the OP wants those.

What about log(x-1)/(x-1) + i pi?

EDIT: Never mind this. It assumed f'(1)=+inf, not f'(0)=+inf.

x^(1/3) - (x-1)^(1/3) + 1 should work though.

Last edited:
a bit late, but thanks all for your efforts!

Char Limit's thing works perfectly! :-) ... after changing the signs:

-x^(1/3)+(x-1)^(1/3)+1

f(x) = sqrt(x) / (1+sqrt(x))
more general : f(x) = (x^a)/(1+(x^a)) with 0<a<1

hmmm, at first i thought -x^(1/3)+(x-1)^(1/3)+1 worked (i tested only for integers) but f is not monotonically increasing (and f' is not monotically decreasing)... and worse, f even gets negative...

but jacquelin your solutions does work perfectly! :-) I am very happy with your quick response! thanks alot!

1. What does "f(0)=0" mean in this context?

This means that when the independent variable, x, is equal to 0, the function has a value of 0. In other words, the point (0,0) lies on the graph of the function.

2. How is it possible for f'(0) to equal positive infinity?

This indicates that the function has a vertical tangent line at the point (0,0). This means that the function is increasing at an infinite rate at this point.

3. What does "f(inf)=<1" tell us about the function?

This means that as the independent variable, x, approaches positive infinity, the function has a maximum value of 1. In other words, the function approaches but never exceeds 1 as x gets larger and larger.

4. Can you provide an example of a function that satisfies these conditions?

One example of a function that satisfies these conditions is f(x) = tan(x), where x is measured in radians. This function has a graph that approaches but never exceeds 1 as x approaches positive infinity, has a vertical tangent line at x=0, and passes through the point (0,0).

5. What type of function would have these properties?

A function that has these properties would likely be a trigonometric function, such as sine, cosine, or tangent, because these functions have periodic behavior and can approach but never exceed certain values as x gets larger. It could also be a logarithmic function, such as ln(x), which has a vertical asymptote at x=0 and approaches but never exceeds 0 as x gets larger.

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