the function below represents the number of people who have a rumor t hours after it has started…

Explanation:

Step1: Find $N(0)$

Substitute $t = 0$ into $N(t)$: [ \begin{align*} N(0)&=\frac{600}{1 + 599e^{-0.5\times0}}\ &=\frac{600}{1+599\times1}\ &=\frac{600}{600}\ & = 1 \end{align*} ] This represents the number of people who start the rumor (the initial - spreaders).

Step2: Find $N(4)$

Substitute $t = 4$ into $N(t)$: [ \begin{align*} N(4)&=\frac{600}{1 + 599e^{-0.5\times4}}\ &=\frac{600}{1+599e^{- 2}}\ &=\frac{600}{1 + 599\times\frac{1}{e^{2}}}\ &=\frac{600}{1+\frac{599}{7.389}}\ &=\frac{600}{1 + 81.07}\ &=\frac{600}{82.07}\approx7 \end{align*} ]

Step3: Find $N(12)$

Substitute $t = 12$ into $N(t)$: [ \begin{align*} N(12)&=\frac{600}{1 + 599e^{-0.5\times12}}\ &=\frac{600}{1+599e^{-6}}\ &=\frac{600}{1 + 599\times\frac{1}{e^{6}}}\ &=\frac{600}{1+\frac{599}{403.43}}\ &=\frac{600}{1 + 1.485}\ &=\frac{600}{2.485}\approx241 \end{align*} ]

Step4: Find when half - the people have heard the rumor

Half of the people is $\frac{600}{2}=300$. Set $N(t)=300$ and solve for $t$: [ \begin{align*} 300&=\frac{600}{1 + 599e^{-0.5t}}\ 1 + 599e^{-0.5t}&=\frac{600}{300}=2\ 599e^{-0.5t}&=1\ e^{-0.5t}&=\frac{1}{599}\ -0.5t&=\ln(\frac{1}{599})=-\ln(599)\ t&=\frac{\ln(599)}{0.5}\approx\frac{6.39}{0.5}=12.78\approx13 \end{align*} ]

Step5: Find when the rumor is spreading fastest

The function $N(t)=\frac{L}{1 + ae^{-bt}}$ (in our case $L = 600,a = 599,b = 0.5$) has its maximum rate of change at $t=\frac{\ln(a)}{b}$. [ \begin{align*} t&=\frac{\ln(599)}{0.5}\ &\approx\frac{6.39}{0.5}=12.78\approx13 \end{align*} ]

Answer:

(a) 1 (b) 7, 241 (c) 13 (d) 13