Question $1[1$ point$]$

The two comportment drug decay model preverted in class anaumes that repeated

drug doaes interact in a linear way, a converience to help present same of the main

dean. Unter this asumption, a person takrx one $11mg$ chew$-$able aspin at the

beginning of every $24$ hour time perid, intendiru to achieve an eventual maximum

nesibual drug level of at iest $15g$ g af aspirin in their blosed plasma $-$ necmary is

treat heart divease. But the persan atops atter the tourth das. See the piece$-$wise

defined function and graph below, deacr hing blood plasma leveh as a function of

time $($s$).$ Cumpute bath the masimum value af aspirin in this peran's blood plaarna

on the interval $[24,481],$ and the time at which it occurs, AND, what are the valuen of

$k_1$ and $k_2$ rapectively? REE AD THE ANS'WERS vERY CARE fULY hERSE.

$10log(\frac{2(1+lo{n}^4)}{1+lo{n}^4}),\frac{81(1+lo{n}^4{)}^2}{2(1+lo{n}^4)},0\mathrm{.}2,0\mathrm{.}1,($bogmans log ban: $10)$

$\frac{81(1+e^{24}{)}^2}{2(1+e^{45})},10ln(\frac{2+2e^{43}}{1+e^{2\mathrm{.}4}}),0\mathrm{.}2,0\mathrm{.}1$

$10ln(\frac{2+2e^{43}}{1+e^{2x}})*\frac{81(1+e^{24}{)}^2}{2(1+e^{4\mathrm{.}5})},0\mathrm{.}1,0\mathrm{.}2$

The arrwer he not in the lat.

$\{\begin{array}{c}\text{:}\frac{81(1+e^{24}{)}^2}{2(1+e^{45})}\\ 10ln(\frac{2+2e^{45}}{1+e^{24}})\\ 0\mathrm{.}1\\ 0\mathrm{.}2\end{array}$

$10ln(\frac{2+2e^{43}}{1+e^{24}})*\frac{81(1+e^{24}{)}^2}{2(1+e^{4\mathrm{.}5})},0\mathrm{.}2,0\mathrm{.}1$can you help solve this