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: $$C(t) = -e^{-3t} + 4.5t^2 + 54$$, where $0 \leq t \leq 9$, $t \in \mathbb{R}$ - Leaving Cert Mathematics - Question 8 - 2018

Substituting t values into the function:

• At $t=0$, $C(0) = 54$.
• At $t=1$, $C(1) = -e^{-3(1)} + 4.5(1)^2 + 54 \approx 57.5$.
• At $t=2$, $C(2) = -e^{-6} + 4.5(4) + 54 \approx 118$.
• At $t=3$, $C(3) = -e^{-9} + 4.5(9) + 54 \approx 175.5$.
• At $t=4$, $C(4) = 224$ (calculated earlier).
• At $t=5$, etc. This pattern continues up to $t=9$.

Plot the calculated values in the intervals provided and connect them smoothly, observing the shape of the function which might illustrate a maximum point and a decrease thereafter.

From the graph at approximately $t = 1.67$, the amount of drug can be estimated by locating the intersection point on the curve.

Estimate this by finding the intersection of the horizontal line $y = 100$ with the curve. It appears to occur shortly after 2 hours.

Differentiate the function:

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

Combine the simplifications to get the rate of change.

Substitute $t=4$ into the derivative: $C'(4) = 3e^{-12} + 36 \approx 42$

Thus, the rate of change after 4 hours is 42 units/hour.

Set the derivative equal to zero and solve for $t$ to find stationary points, then evaluate $C(t)$ at $t=6$ to find the maximum.

Evaluate $C'(7)$; if it is negative, this indicates the drug amount is decreasing at that point in time.