Answer
The first inflection point occurs at $t = 28.6~minutes$. This is the time when the level of medication is increasing at its highest rate.
The second inflection point occurs at $t = 85.7~minutes$. This is the time when the level of medication is decreasing at its highest rate.
Work Step by Step
$S(t) = At^pe^{-kt}$
We can express the equation with the values given in the question:
$S(t) = 0.01~t^4~e^{-0.07~t}$
We can find $S'(t)$:
$S'(t) = 0.01~(4t^3~e^{-0.07~t}-0.07~t^4~e^{-0.07~t})$
We can find $S''(t)$:
$S''(t) = 0.01~(12t^2~e^{-0.07~t}-(0.07)~4t^3~e^{-0.07~t}-(0.07)~4t^3~e^{-0.07~t}+(0.07)^2~t^4~e^{-0.07~t})$
$S''(t) = 0.01~(12t^2~e^{-0.07~t}-0.56~t^3~e^{-0.07~t}+0.0049~t^4~e^{-0.07~t})$
We can find the values of $t$ where there are inflection points:
$S''(t) = 0$
$S''(t) = 0.01~(12t^2~e^{-0.07~t}-0.56~t^3~e^{-0.07~t}+0.0049~t^4~e^{-0.07~t}) = 0$
$t^2~e^{-0.07~t}~(12-0.56~t+0.0049~t^2) = 0$
$t = 0~~$ or $~~0.0049~t^2-0.56~t+12 = 0$
We can use the quadratic formula:
$t = \frac{0.56\pm \sqrt{(-0.56)^2-(4)(0.0049)(12)}}{2(0.0049)}$
$t = \frac{0.56\pm \sqrt{0.0784}}{0.0098}$
$t = 28.6, 85.7$
The first inflection point occurs at $t = 28.6~minutes$. This is the time when the level of medication is increasing at its highest rate.
The second inflection point occurs at $t = 85.7~minutes$. This is the time when the level of medication is decreasing at its highest rate.
