Answer
choice (C)
Work Step by Step
Equation form:
$\displaystyle \quad f(x)=\frac{N}{1+Ab^{-x}}$,
where $N$ = limiting value,
All three need to be divided (up and down) with 2:
$A:\displaystyle \quad \frac{9}{1+3.5(5)^{-x}}$
$B:\quad \frac{9}{1+1.5(1.1)^{-x}}$
$C:\quad \frac{9}{1+3.5(1.1)^{-x}} $
Reading the graph, $N=9,$ so no choice is eliminated.
$f(0)=2\Rightarrow\left\{\begin{array}{ll}
2 & =\frac{9}{1+Ab^{0}}\\
1+A & =9/2\\
A & =4.5-1=3.5
\end{array}\right.\qquad\Rightarrow\qquad $(B) is eliminated
For small x, $f(x)\displaystyle \approx(\frac{N}{1+A})b^{x}=\frac{9}{4.5}b^{x}=2b^{x}$
The logistic function grows approximately exponentially with base b.
The base is either $1.1$ or $5.$
If it is 5, $f(x+1)=5\cdot f(x),$
a change in x from $0$ to $1$ should change $f(x)$ from $2$ to $10,$ and it does not, so $b$ must be smaller than $5$. Thus, we have: choice (C)
