Question

students in a psychology class took a final examination. as part of an experiment to see how much of the course content they remembered over time, they took equivalent forms of the exam in monthly intervals thereafter. the average score for the group f(t), after t months was modeled by the function f(t)=90 - 14 ln(t + 1), 0 ≤ t ≤ 12. a. what was the average score on the original exam? b. what was the average score after 8 months? c. sketch the graph of f (either by hand or with a graphing utility.) describe what the graph indicates in terms of the material retained by the students.

Explanation:

Step1: Find the average score on the original exam (t = 0)

Substitute \(t = 0\) into \(f(t)=90 - 14\ln(t + 1)\)

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Step2: Find the average score after 8 months (t = 8)

Substitute \(t = 8\) into \(f(t)=90 - 14\ln(t + 1)\)

$$
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$$

Answer:

a. The average score on the original exam is \(90\).
b. The average score after 8 months is approximately \(59.24\).
c. To sketch the graph of \(y = f(t)=90 - 14\ln(t + 1)\), we know that the domain is \(0\leq t\leq12\). The \(y\) - intercept is at \((0,90)\). As \(t\) increases, the function \(y = f(t)\) is a decreasing function because the derivative \(f^\prime(t)=-\frac{14}{t + 1}<0\) for \(t\geq0\). The graph indicates that as time (\(t\)) passes (from \(t = 0\) to \(t=12\)), the average score of the students (which represents the material retained) decreases. The rate of decrease slows down as \(t\) gets larger (since the function is concave - up, \(f^{\prime\prime}(t)=\frac{14}{(t + 1)^2}>0\) for \(t\geq0\)).