AP Precalculus -2.14 Logarithmic Function Modeling- FRQ Exam Style Questions - Effective Fall 2023

(i) Writing the equations:
We are given the following data points:
• At \(t=2\), \(R(2) = 15\).
• At \(t=6\), \(R(6) = 67\).

For \(t=2\), since \(0 \le 2 < 6\), we use the first part of the piecewise function: \(R(t) = 7(a)^{t/2}\).
$$15 = 7(a)^{2/2} \quad \Rightarrow \quad 15 = 7a^1$$
Equation 1: \(15 = 7a\)

For \(t=6\), since \(t \ge 6\), we use the second part of the piecewise function: \(R(t) = -213.29 + b \ln t\).
Equation 2: \(67 = -213.29 + b \ln(6)\)

(ii) Finding the values for \(a\) and \(b\):
From Equation 1:
$$a = \frac{15}{7} \approx 2.1428$$
From Equation 2:
$$67 + 213.29 = b \ln(6)$$
$$280.29 = b \ln(6)$$
$$b = \frac{280.29}{\ln(6)} \approx \frac{280.29}{1.79176} \approx 156.4328$$
Answer: \(a \approx 2.143\), \(b \approx 156.433\)

(i) Average Rate of Change:
The formula for the average rate of change from \(t=2\) to \(t=6\) is:
$$\text{Avg Rate} = \frac{R(6) – R(2)}{6 – 2}$$
Substituting the given values (\(R(6)=67\) and \(R(2)=15\)):
$$\text{Avg Rate} = \frac{67 – 15}{4} = \frac{52}{4} = 13$$
Answer: 13 students per hour.

(ii) Interpretation:
On average, the number of students who have heard the rumor increases by 13 students per hour between the 2nd hour and the 6th hour.

(iii) Estimates vs. Model Prediction:
Answer: The estimates are greater than the number of students predicted by the model.
Reasoning:
• On the interval \(0 < t < 6\), the function \(R(t) = 7(a)^{t/2}\) is an exponential growth function with a base greater than 1.
• Exponential growth functions are concave up (the rate of change is increasing).
• The average rate of change corresponds to the slope of the secant line connecting the points at \(t=2\) and \(t=6\).
• For a concave up curve, the secant line lies above the curve on the interval between the two points. Therefore, linear estimates based on the average rate (secant line) will be greater than the actual function values.