I need to find a mathematical function that looks like this curve:
Please look at this picture:
- $f_{max}(x)= 10$
- $f(x_{end}) = 5$
- $f(0) = 0$
I already have this function:
$f(x) = a*x^2*e^{b*x^2}+c$
But this function has an asymptote that goes to 0:
$\lim \limits_{x \to \infty}\rightarrow 0$
But my goal is, that the function has a finite final value that is greater than 0 and is not an asymptote. Is that possible?
Thanks in advance!
1) You can try this function $f(x) = \frac{Ax^2 + Bx +C}{Dx^2+Ex+F}$.
Condition $f(x)=0$ implies that $C=0$. Hence, $f(x) = \frac{Ax^2 + Bx}{Dx^2+Ex+F}$.
Condition $\lim_{x\to \infty}f(x)=5$ implies that $A=5D$. We can also choose $D=1$ for simplicity now, but you can pick $D$ at your convenience. Then $$f(x)=\frac{5x^2 + Bx}{x^2+Ex+F}.$$
It remains to impose that $f'(a)=0$ at some point $x=a$ and $f'(x)>0$ for all $x\in [0,a)$ and $f'(x)<0$ for all $x\in (a,\infty)$. Also, $f(a)=10$.
First, $f(a)=\frac{5a^2 + Ba}{a^2+Ea+F}=10$ implies $10a^2 + (10E-B)a + 10F=0$.
Also, $$f'(x) = \frac{(5E-B)x^2+10Fx+BF}{(x^2+Ex+F)^2}.$$ The critical points are: $$x_{crit}= \frac{-10F \pm \sqrt{100F^2-4BF(5E-B)}}{2(5E-B)}.$$ So, you need $100F^2 > 4BF(5E-B)$ and then $a$ is the positive root.
I think in particular if you choose $B>E$ and $a=\frac{-10F - \sqrt{100F^2-4BF(5E-B)}}{2(5E-B)}$ you get one possible choice.
For example, $A=5$, $B=2$, $C=0$, $D=1$, $E=1$ and $F$ as you want. $$f(x)=\frac{5x^2 + 2x }{x^2+x+F}$$ you get a function with your properties. Another question is if you can tune the parameters $A$, $B$, $C$, $D$, $E$ and $F$ to make the function have more prominent maximum and decreasing shape towards $y=5$.
2) So, another choice is to simply take: $$f(x) = \begin{cases} Ax^2+Bx+C, \quad x\in [0,k),\\ D, \quad x\in [k,\infty), \end{cases} $$ for some constants $A$, $B$, $C$ and $k$.
If $A<0$ then the function $f(x) = Ax^2+Bx$ is a family of concave parabolas passing through the points $(0,0)$ and $(-\frac{B}{A},0)$ and its maximum is at point $(-\frac{B}{2A},-\frac{B^2}{4A})$. Since $-\frac{B^2}{4A}=10$ we have $B^2 = -40A$. Now, you want a line on $y=5$. The intersection of $y=5$ with the parabola are the two points: $x_{1,2} = \frac{-B\pm \sqrt{B^2 - 20A}}{2A}$. The right point is $$k=\frac{-B- \sqrt{B^2 - 20A}}{2A}$$ and you can thus define
$$f_{A,B}(x) = \begin{cases} Ax^2+Bx, \quad x\in [0,k),\\ 5, \quad x\in [k,\infty), \end{cases}$$
where
$$k=\frac{-B- \sqrt{B^2 - 20A}}{2A}$$ for all $A$ and $B$ such that $A<0$ and $-\frac{B^2}{4A}=10$.
The family of functions $f_{A,B}$ have a maximum at $x=a=-\frac{B}{A}$ and such that $f_{A,B}(a)=10$ and $\lim_{x\to \infty}f_{A,B}(x)=5$.
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