Question

the population growth of a specific city (in thousands of people), t years after it was founded, is modelled by the following equation.
$f(t) = 22\left(1 + e^{12 - t}\
ight)^{-1}$
what is the instantaneous rate of change of the population, 6 years after the town was founded?
show your work here
hint: to add an exponent $(x^y)$, type \exponent\ or press \^\

Explanation:

Step1: Rewrite the function

Rewrite $f(t)$ as $f(t) = \frac{22}{1 + e^{12-t}}$

Step2: Apply quotient rule for derivative

Let $u=22$, $v=1+e^{12-t}$. The quotient rule is $f'(t)=\frac{u'v - uv'}{v^2}$.
$u'=0$, $v' = -e^{12-t}$
$f'(t)=\frac{0 \cdot (1+e^{12-t}) - 22 \cdot (-e^{12-t})}{(1+e^{12-t})^2} = \frac{22e^{12-t}}{(1+e^{12-t})^2}$

Step3: Substitute $t=6$ into $f'(t)$

First calculate $12-t=12-6=6$
$f'(6)=\frac{22e^{6}}{(1+e^{6})^2}$
We can simplify $e^{12-6}=e^6$, so:
$f'(6)=\frac{22e^{6}}{(1+e^{6})^2}$
We can also rewrite this as $\frac{22}{(e^{-3}+e^{3})^2}$ but the expanded form is also valid. Calculating the numerical value:
$e^6\approx403.4288$, so $1+e^6\approx404.4288$, $(1+e^6)^2\approx163562.6$, $22e^6\approx8875.4336$
$f'(6)\approx\frac{8875.4336}{163562.6}\approx0.0543$

Answer:

The instantaneous rate of change is $\frac{22e^{6}}{(1+e^{6})^2}$ (or approximately 0.0543 thousand people per year, i.e., 54.3 people per year)