In search of a formula which fullfills criteria

Question

I've got a few criteria and I need to find a simple (as simple as possible) function which fits those criteria:

$$\lim_{x \to -\infty }{f(x)} = C_1$$ $$f(0) = C_2$$ $$f'(0) = C_3$$

Where $C_1,C_2,C_3$ are constants... ($C_1$ will be $1$, and the other constants are determined from another function so the function is continuous).

The function only has to be defined to $x = 0$. (As for the positive $x$ another function would be used).

I'm thinking along the line of:

$$f(x) = \frac{ 1 }{ax + b} + c$$

However I'm more or less struck on "using" the limit constraint to solve this problem.

$$f(0) = \frac{1}{b} + c = C_2$$ $$f'(0) = - \frac{a}{b^2} = C_3$$

Thanks in advance, paul23

Answer 1

If you choose, $f(x) = c + \frac{1}{ax+b}$.

$$C_1 = \lim_{x \rightarrow -\infty} f(x) = c$$ provided $a \neq 0$. Hence $c = C_1$.

$$C_2 = f(0) = C_1 + \frac{1}{b} \Rightarrow b = \frac{1}{C_2-C_1}$$

$$C_3 = f'(0) = -\frac{a}{b^2} \Rightarrow a = -\frac{C_3}{(C_2-C_1)^2}$$

Hence, $a = -\frac{C_3}{(C_2-C_1)^2}$, $b = \frac{1}{C_2-C_1}$ and $c=C_1$.

$$f(x) = \frac{(C_2-C_1)^2}{-C_3 x + (C_2 - C_1)} + C_1$$

An other possible candidate is $$f(x) = (C_2-C_1) e^{\frac{C_3}{C_2-C_1}x} + C_1$$ provided $C_3 \times(C_2 - C_1) > 0$.

Answer 2

If you define $f(x)=\frac{(C_1-C_2)x}{x+a}+C_2$ you get the proper limit at $-\infty$ and value at 0. So then $f'(0)=-\frac{(C_1-C_2)a}{a^2}=C_3$ or $a=\frac{-(C_1-C_2)}{C_3}$

Note: parentheses added in the last after original posting