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:
$\quad A:\displaystyle \quad \frac{7}{1+2.5(15)^{-x}}$
$\quad B:\displaystyle \quad \frac{7}{1+7.5(1.05)^{-x}}$
$\quad C:\displaystyle \quad \frac{7}{1+2.5(1.05)^{-x}} $
Reading the graph, $N=$7, so no choice is eliminated.
$f(0)=2\Rightarrow\left\{\begin{array}{ll}
2 & =\frac{7}{1+Ab^{0}}\\
1+A & =7/2\\
A & =3.5-1=2.5
\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 $1.05$ or $15.$
If it is $15$, then $f(x+1)=15(f(x))$. A change in x from $0$ to $1$ should change $f(x)$ from $2$ to $30,$ approximately and it does not,
so $b$ must be smaller than $15$. Hence, we have: choice ($C$).
