The amount, in appropriate units, of a certain medicinal drug in the bloodstream t hours after it has been taken can be estimated by the function: $C(t) = -3e^{-t} + 4.5t^2 + 54$, where $0 \leq t \leq 9$, $t \in \mathbb{R}$ - Leaving Cert Mathematics - Question 8 - 2018

(Values calculated by substituting each $t$ into the function $C(t)$).

The graph is plotted based on the values in the completed table, showing the relationship between time and the amount of the drug in the bloodstream. Ensure proper labeling of axes and points. The curve should reflect the drug amount peaking around $t=4$ before gradually declining.

i) The amount of the drug in the bloodstream after $\frac{3}{2}$ hours:

Estimate from the graph, which is approximately 140 units.

ii) How long after taking the drug will the amount of the drug be 100 units?

Estimate from the graph, which is roughly 1 hour and 45 minutes.

To find the rate of change of the drug amount, we differentiate $C(t)$ with respect to $t$:

$C'(t) = -3e^{-t} + 9t + 54$

This gives us the expression for the rate at which the drug amount is changing.

Substituting $t = 4$ into $C'(t)$:

$C'(4) = -3e^{-4} + 9(4) + 54 \approx -0.198 + 36 + 54 \approx 89.802 \text{ units/hour}$

To find the maximum, evaluate $C(t)$ at critical points and endpoints between $0$ and $9$ hours. The max occurs at $t = 6$, where:

$C(6) = -3(6^2) + 4.5(6^2) + 54 = 270 \text{ units}$

Substituting $t = 7$ into $C'(t)$:

$C'(7) = -3e^{-7} + 9(7) + 54 \approx -0.002 + 63 + 54 = 117 \text{ units/hour}$

Since $C'(t)$ is positive at $t = 7$, the drug amount is still increasing, not decreasing.