How to find the point of symmetry for $f(x) = \frac{6}{1+3\cdot(0.4)^x}$ algebraically? In analyzing the graph of the function $$f(x) = \frac{6}{1+3\cdot(0.4)^x}$$ the textbook lists that it has symmetry around the point (1.2, 3). Is there a way to find the coordinates of this point algebraically, without tracing the graph or using calculus to find the inflection point?
Thank you!
 A: The value of the function at $-\infty$ is $0$, and at $+\infty$ is $6$ ( horizontal asymptotes if one is precise). Now if the graph has a center of symmetry, it is at the point of height $\frac{6}{2}=3$. So let $x_0$ such that $f(x_0) = 3$. This is equivalent to $3 (0.4)^{x_0}=1$.
One should check that $f(x_0- h) + f(x_0+ h) = 2 f(x_0)$.
This is equivalent to
$$\frac{6}{1 + 0.4^{-h}} + \frac{6}{1 + 0.4^h} = 6$$
or, with $0.4^h = t$
$$\frac{1}{1+1/t} + \frac{1}{1+t} = 1$$
which is true.
Note: $f$ is a logistic function.
A: It is so simple with calculus
$$f(x)=\frac{a}{1+b c^x}$$ $$f'(x)=-\frac{a b c^x \log (c)}{\left(b c^x+1\right)^2}\qquad \text{and} \qquad f''(x)=\frac{a b c^x \log ^2(c) \left(b c^x-1\right)}{\left(1+b c^x\right)^3}$$
So, the inflection point is at
$$x_*=-\frac{\log (b)}{\log (c)}\implies f(x_*)=\frac a 2$$
This was the long way.
The short way is to look for $x$ such that
$$\frac{a}{1+b c^x}=\frac a 2\implies 1+b c^x=2\implies b c^x=1\implies x=-\frac{\log (b)}{\log (c)}$$
