Problem 82 - Contemporary Precalculus

Question

Students in a precalculus class were given a final exam. Each month thereafter, they took an equivalent exam. The class average on the exam taken after \(t\) months is given by $$ F(t)=82-8 \cdot \ln (t+1) $$ (a) What was the class average after six months? (b) After a year? (c) When did the class average drop below \(55 ?\)

Step-by-Step Solution

Answer

(a) After six months, the class average is approximately 69.38. (b) After a year (12 months), the class average is approximately 60.17. (c) The class average drops below 55 after approximately 28.14 months.

Step 1: (a) Find the class average after six months.

The given function is $$ F(t) = 82 - 8 \cdot \ln(t + 1) $$ To find the class average after six months, we need to find \(F(6)\): $$ F(6) = 82 - 8 \cdot \ln(6 + 1) $$ $$ F(6) = 82 - 8 \cdot \ln(7) $$ Now, use a calculator to compute the value of \(F(6)\): $$ F(6) \approx 69.38 $$ So, the class average after six months is approximately \(69.38\).

Step 2: (b) Find the class average after a year.

To find the class average after a year (12 months), we need to find \(F(12)\): $$ F(12) = 82 - 8 \cdot \ln(12 + 1) $$ $$ F(12) = 82 - 8 \cdot \ln(13) $$ Now, use a calculator to compute the value of \(F(12)\): $$ F(12) \approx 60.17 $$ So, the class average after a year is approximately \(60.17\).

Step 3: (c) Determine when the class average drops below 55.

We will solve the inequality: $$ F(t) < 55 $$ This inequality is equivalent to: $$ 82 - 8 \cdot \ln(t + 1) < 55 $$ Now, subtract 82 from both sides to isolate the logarithm term: $$ -8 \cdot \ln(t + 1) < -27 $$ Then divide both sides by -8 (remember to switch the inequality sign when dividing by a negative number): $$ \ln(t + 1) > 3.375 $$ Now, to eliminate the natural logarithm, we will take the exponent of both sides with base \(e\): $$ e^{\ln(t + 1)} > e^{3.375} $$ Simplifying, we get: $$ t + 1 > 29.14 $$ Therefore, $$ t > 28.14 $$ The class average drops below 55 after approximately 28.14 months.

Key Concepts

Logarithmic Functions

Logarithmic functions are crucial in precalculus, especially when analyzing models like exponential growth and decay. They help convert complex multiplicative relationships into simpler additive ones, making them easier to handle and solve.
The function in our exercise, \(F(t) = 82 - 8 \cdot \ln(t+1)\), neatly illustrates how logs are used to model decay over time. Here, the term \(\ln(t+1)\) (natural logarithm) plays a key role in understanding the rate at which the class average decreases.

• The base of the natural log, \(e\), is approximately equal to 2.718. The natural log can be used to invert exponential functions.
• It helps determine how fast a quantity is changing, essential for calculating averages or decay over time.
• When computing \(F(t)\), the logarithmic function affects the deduction (subtraction) from the constant 82, showcasing the impact of intervals \((t+1)\) on class averages.

Understanding this setup allows students to analyze the trend accurately, predicting values through simple calculation.

Inequalities in Precalculus

Inequalities are fundamental in solving problems where you need to determine if one side of the equation is greater or less than the other. In the exercise, inequalities help us find when the class average drops below a certain score, \(55\).

This involves transforming the inequality \(F(t) < 55\) using logs.

• First, we rewrite \(82 - 8 \cdot \ln(t + 1) < 55\) for easier manipulation. Subtracting 82 helps spotlight the log term.
• Dividing by a negative number requires flipping the inequality sign, illustrating a critical property of inequalities.
• Taking the exponential \(e\) of each side effectively "undoes" the log, providing a solvable expression \(t + 1 > 29.14\).

This shows the timing around when the class average would fall below the threshold, reinforcing practical use in real-world scenarios.

Function Evaluation

Function evaluation is straightforward yet vital in precalculus. It involves substituting specific values into a function to determine outputs. In this exercise, \(F(t) = 82 - 8\cdot \ln(t+1)\) is evaluated for different months.
For instance:

• When examining \(F(6)\), inputting 6 in place of \(t\) provides the average after half a year. The function is solved numerically to yield \(F(6) \approx 69.38\).
• Similarly, plugging in 12 calculates the average for a full year, resulting in \(F(12) \approx 60.17\).

Evaluation like this helps students visualize changes over time and supports understanding of function behavior. Mastery of function evaluation means students can easily analyze mathematical models and predict outcomes.