Problem 54 - A Graphical Approach to Precalculus with Limits

Question

The average cost of a 30 -second Super Bowl ad in millions of dollars is approximated by the piece wise-defined function $$f(x)=\left\\{\begin{array}{ll} 0.0475 x-93.3 & \text { if } 1967 \leq x \leq 1998 \\ 0.179 x-356.037 & \text { if } 1998< x \leq 2017 \end{array}\right.$$ where \(x\) represents the year from 1967 to 2017 . Find and interpret each function value. Round values to the nearest hundredth of a million dollars. (b) \(f(1998)\) (a) \(f(1967)\) (c) \(f(2017)\) (d) Is \(f\) continuous on its domain? (e) Graph \(f\).

Step-by-Step Solution

Answer

(a) 0.06 million, (b) 1.61 million, (c) 4.01 million, (d) Not continuous.

Step 1: Evaluate f(1998)

For the year 1998, we will use the first piece of the piecewise function because it is valid for the interval \(1967 \leq x \leq 1998\). The function is given by:\[ f(x) = 0.0475x - 93.3 \]Plug in \(x = 1998\):\[ f(1998) = 0.0475 \times 1998 - 93.3 \]Calculate the product:\[ = 94.905 - 93.3 \]Subtract:\[ = 1.605 \]Therefore, \(f(1998) = 1.61\) million dollars.

Step 2: Evaluate f(1967)

For the year 1967, we use the first piece of the piecewise function as it applies to \(1967 \leq x \leq 1998\).The function is:\[ f(x) = 0.0475x - 93.3 \]Plug in \(x = 1967\):\[ f(1967) = 0.0475 \times 1967 - 93.3 \]Calculate the product:\[ = 93.3575 - 93.3 \]Subtract:\[ \approx 0.0575 \]Rounded to the nearest hundredth, \(f(1967) = 0.06\) million dollars.

Step 3: Evaluate f(2017)

For the year 2017, we will use the second piece of the function since it is valid for \(1998 < x \leq 2017\).The function is:\[ f(x) = 0.179x - 356.037 \]Plug in \(x = 2017\):\[ f(2017) = 0.179 \times 2017 - 356.037 \]Calculate the product:\[ = 360.043 - 356.037 \]Subtract:\[ = 4.006 \]Rounded to the nearest hundredth, \(f(2017) = 4.01\) million dollars.

Step 4: Analyze Continuity

To determine if \(f\) is continuous over its domain, we need to evaluate the continuity at \(x = 1998\), where the function pieces meet.Calculate \(\lim_{x \to 1998^-} f(x)\) using the first piece:\[ = 0.0475 \times 1998 - 93.3 \approx 1.61 \]Calculate \(\lim_{x \to 1998^+} f(x)\) using the second piece:\[ = 0.179 \times 1998 - 356.037 \approx 0.375 \]Since the limits are not equal (\(1.61 eq 0.375\)), the function \(f\) is not continuous at \(x = 1998\), and thus not continuous over the entire domain.

Step 5: Graph the Function

Graphing involves plotting the two linear pieces over their respective intervals. 1. Plot the first piece from $x = 1967$ to $x = 1998$: It starts near $0.06$ and ends at $1.61$. 2. Plot the second piece from $x = 1998$ to $x = 2017$: It starts near $0.375$ and ends at $4.01$. Join each linear segment on its respective interval, respecting their calculated starting and finishing values.

Key Concepts

Continuity

In mathematics, a function is said to be continuous over a domain if there are no breaks, jumps, or holes in its graph within that domain. For piecewise functions, such as the one given in this exercise, continuity is evaluated at the points where the function changes its rule — typically at the endpoints of each piece.
Let's analyze the piecewise function \(f\) for continuity. This function splits at \(x = 1998\). To determine continuity at this point, we calculate the approaches from both sides:

• Using the first piece, the limit as \(x\) approaches 1998 from the left is approximately \(1.61\).
• Using the second piece, the limit as \(x\) approaches 1998 from the right is approximately \(0.375\).

As these limits are not equal, the function \(f\) is not continuous at \(x = 1998\). Hence, \(f\) is not continuous over its entire domain, which means its graph will have a noticeable jump at \(x = 1998\). Understanding continuity helps predict the behavior of functions and their graphs.

Graphing

Graphing a piecewise function involves plotting each segment of the function based on its equation. Each piece of the function has a specific interval for which it is valid. Here’s how you would graph the given piecewise function.
For the first segment, the equation is \(f(x) = 0.0475x - 93.3\). This part of the function is active from the year 1967 to 1998. To graph this:

• Calculate the function values at the endpoints: \(f(1967) \approx 0.06\) and \(f(1998) \approx 1.61\).
• Plot these points on the graph and draw a line connecting them.

Next, the second part of the function is \(f(x) = 0.179x - 356.037\), effective from 1998 to 2017:

• Again, calculate the function values at the endpoints: \(f(1998) \approx 0.375\) and \(f(2017) \approx 4.01\).
• Plot these points on the graph and draw a line connecting them.

When graphing, note the gap at \(x = 1998\) due to discontinuity. The first segment ends at \(f(1998) = 1.61\) and the second begins at \(f(1998) = 0.375\). This visual representation helps in understanding the piecewise nature of the function and any discontinuities.

Function Evaluation

Evaluating a piecewise function means determining the function's output for specific input values, considering which equation is valid.
For the exercise given:

• Calculate \(f(1998)\): The year's within the first piece of the function, \(f(x) = 0.0475x - 93.3\). Plug in \(x = 1998\) to get \(f(1998) \approx 1.61\) million dollars.
• Calculate \(f(1967)\): Similarly, for the beginning of the first segment, calculate \(f(1967) \approx 0.06\) million dollars using the respective piece.
• Calculate \(f(2017)\): For this recent year, use the second piece, \(f(x) = 0.179x - 356.037\), obtaining \(f(2017) \approx 4.01\) million dollars.

Function evaluation is crucial for understanding what the function represents at particular points and can highlight changes between periods or intervals. In this case, it helps us see how the cost of Super Bowl ads has evolved by year within the prescribed intervals.