After a long period of time everyone knows the rumour!

\begin{align*} lim_{t \to \infty}\fe{p}{t}&=\frac{1}{1+ a e^{-k\infty}}\\ \text{Since }e^{-k\infty} = 0, \{ k \in \mathbb{R} | k > 0 \}\\ \therefore lim_{t \to \infty} p(t)&= \frac{1}{1+0}\\ &=1 \end{align*}

\begin{align*} u&= 1 + a e^{-kt}\\ \frac{dp}{dt}&=\frac{dp}{du}\frac{du}{dt}&\text{(Chain Rule)}\\ &=\frac{-1}{u^2}\left(-ak e^{-kt}\right)\\ &=\frac{ak e^{-kt}}{(1+a e^{-kt})^2} \end{align*}

\begin{align*} p&= \frac{1}{1+ a e^{-kt}}\\ \text{Substitute p's for t's, and vice versa:}\\ t&= \frac{1}{1+ a e^{-kp}}\\ \text{Solve for p:}\\ 1 + a e^{-kp}&=\frac{1}{t}\\ a e^{-kp}&=\frac{1}{t} - 1\\ e^{-kp}&=\frac{1}{at} - \frac{1}{a}\\ -kp&=\ln \left(\frac{1}{at} - \frac{1}{a}\right)\\ p&=-\frac{1}{k}\ln \left(\frac{1-t}{at}\right)\\ p&=\frac{1}{k}\ln \left(\frac{at}{1-t}\right)\\ \text{In other words the inverse is:}\\ p^{-1}&=\frac{1}{k}\ln \left(\frac{at}{1-t}\right) \end{align*}

\(7.38\) days.

See Sage plot  of \(p(t)\) below.

Use the inverse function with a value of \(0.8\) \begin{align*} p^{-1}&=\frac{1}{k}\ln \left(\frac{at}{1-t}\right)\\ p^{-1}(0.8)&=7.378 \end{align*}

See Sage plot  of inverse \(p(t)\) below.

The mass (in \(mg\)) after time \(t\) (in days) is: \begin{equation*}M_t=800 e^{-0.1386t}\end{equation*}

The rate of change of mass is proportional to the current mass: \begin{equation*}\frac{dM}{dt} \propto M\end{equation*} Lets call the constant of proportionality \(\lambda\). \begin{equation*}\frac{dM}{dt} = -\lambda M\end{equation*} The negative is because this is a decay.

The solution of this equation is: \begin{equation*}M_t = M_0 e^{-\lambda t}\end{equation*} (where \(M_0\) is the mass at \(t=0\).)

At the half-life \(t=T_{\frac{1}{2}}\) and \(M_t = \frac{M_0}{2}\): \begin{equation*}\frac{M_0}{2} = M_0 \exp\left({-\lambda T_{\frac{1}{2}}}\right)\end{equation*} \begin{equation*}2 = \exp\left({\lambda T_{\frac{1}{2}}}\right)\end{equation*} \begin{equation*}\log{2} = \lambda T_{\frac{1}{2}}\end{equation*} \begin{equation*}\lambda = \frac{log2}{T_{\frac{1}{2}}}\end{equation*}

We are told that the half-life is \(5\) days. \begin{equation*}\lambda \approx 0.1386\end{equation*}

Substituting with the original mass gives us: \begin{equation*}M_t = 800 e^{-0.1386t}\end{equation*} where \(M_t\) is the mass in \(mg\) and \(t\) is the time in days.

\(N(30) = 12.5 mg\)

\begin{equation*}N(30) = 800 e^{-0.1386 \times 30} = 800 e^{-4.1589} = 800 \times 0.015625 = 12.5 \end{equation*}

\(48.2\) days.

\begin{align*} M_t&=M_0 e^{-\lambda t}\\ 1&=800e^{-\lambda t}\\ \frac{1}{800}&=e^{-\lambda t}\\ 800&=e^{\lambda t}\\ log{800}&=\lambda t\\ t&=\frac{log{800}}{\lambda}\\ &=\frac{log{800}}{0.1386}\\ & \approx 48.2 \end{align*}

See Sage plot  of mass function below.

The first derivative is given by: \begin{equation*}\frac{\Delta f(x)}{\Delta x} = \lim_{h \to 0}\frac{f(x+h)-f(x)}{(x+h)-(x)} = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\end{equation*}

The solution is to apply this twice. The trick here is to start with the correct increments (see Math Stack Q&A [3.7.3.1.1]).

\begin{align*} \fe{f''}{x}&=\lim_{h \to 0} \frac{\fe{f'}{x} - \fe{f'}{x-h}}{h}\\ &=\lim_{h \to 0}\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}\\ &=\lim_{h \to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2} \end{align*}

First set \(a=x+h\): \begin{equation*}f(x)f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots\end{equation*} TBD

\begin{align*} \text{Apply the mean value theorem}\\ &f'(c)=\frac{f(8) - f(2)}{8-2}\\ \text{Since we know}\\ &3 \leq \fe{f'}{x} \leq 5\\ \text{We get}\\ &3 \leq \frac{f(8) - f(2)}{6} \leq 5\\ &18 \leq f(8) - f(2) \leq 30 \end{align*}