Function with two conditions

Question

I need a function that fulfills these two conditions:

$$ y'(0) = 1 $$ $$ \lim_{x\to\infty} y(x) = 70 $$

This results in a function that has two tangents: $y=x$ (blue) and $y=70$ (purple)

I have already figured out one function that fulfills both conditions: $y=\frac{70x}{x+70}$ (red)

The problem is that this function is too "slow" for my application. Is there any preferably simple equation that keeps the function's starting slope near 1 for a longer time so it "reaches" its limit (70) quicker? I made a quick drawing of what I mean:

The function will never receive negative $x$ values, so the part left of the y-axis doesn't matter. The function should never cross any of its tangents.

Answer 1

Why don't you use an exponential? $$f(x) = 70(1-e^{-\frac{x}{70}})$$ Then you would have $$f'(x) = \frac{70}{70}e^{-x} = e^{-x} $$ And trivially $f'(0) = 1$. Bessides that $$\lim_{x\to\infty} f(x) = 70\lim_{x\to\infty}(1-e^{-\frac{x}{70}}) = $$ $$70(1-\lim_{x\to\infty}e^{-\frac{x}{70}}) = 70(1-0) = 70 $$ Above this all, $f$ has an exponential aproximation to the assymptote in $y = 70$

Answer 2

How about $$ f(x) = \frac{70 \frac{1}{k} ((x+1)^k - 1)}{\frac{1}{k} ((x+1)^k - 1) + 70} \text{?} $$ Then $$ f'(x) = \frac{4900 k^2 (x+1)^{k-1}}{\left((x+1)^k+70 k-1\right)^2} \text{,} $$ so $$ f'(0) = 1 \text{.} $$

For example, with $k = 2$:

$k$ need not be an integer. For example, $k = \pi$.

Answer 3

Have you considered the arctangent function? We can create a function in the form $$f(x)=a\arctan k(x-c)+d$$ which has a horizontal asymptote at $y=\dfrac{a\pi}{2}+d$.

Its derivative is $$f'(x)=\frac{ak}{1+[k(x-c)]^2}\implies f'(0)=\frac{ak}{1+k^2c^2}=1.$$

Solve for the appropriate parameters that you like.