Problem 25 - Applied Calculus

Question

Determine whether each of the following tables of values could correspond to a linear function, an exponential function, or neither. For each table of values that could correspond to a linear or an exponential function, find a formula for the function. $$ \begin{array}{l} \text { (a) }\\\ \begin{array}{l|l} \hline x & f(x) \\ \hline 0 & 10.5 \\ 1 & 12.7 \\ 2 & 18.9 \\ 3 & 36.7 \\ \hline \end{array} \end{array} $$ $$ \begin{array}{l} \text { (b) }\\\ \begin{array}{c|l} \hline t & s(t) \\ \hline-1 & 50.2 \\ 0 & 30.12 \\ 1 & 18.072 \\ 2 & 10.8432 \\ \hline \end{array} \end{array} $$ $$ \begin{array}{l} \text { (c) }\\\ \begin{array}{c|c} \hline u & g(u) \\ \hline 0 & 27 \\ 2 & 24 \\ 4 & 21 \\ 6 & 18 \\ \hline \end{array} \end{array} $$

Step-by-Step Solution

Answer

(a) Neither; (b) Exponential: \(s(t) = 50.2 \times 0.6^{t+1}\); (c) Linear: \(g(u) = -1.5u + 27\)

Step 1: Examine Linear Function for (a)

To determine if (a) could be a linear function, we calculate the differences between consecutive y-values (f(x)).These differences are:- Between 0 and 1: \(12.7 - 10.5 = 2.2\)- Between 1 and 2: \(18.9 - 12.7 = 6.2\)- Between 2 and 3: \(36.7 - 18.9 = 17.8\)Since the differences are not constant, the function is not linear.

Step 2: Examine Exponential Function for (a)

To check for an exponential function, compute the ratio of consecutive values:- Between 0 and 1: \(\frac{12.7}{10.5} \approx 1.21\)- Between 1 and 2: \(\frac{18.9}{12.7} \approx 1.49\)- Between 2 and 3: \(\frac{36.7}{18.9} \approx 1.94\)Since the ratios of differences are not constant, (a) cannot be modeled as an exponential function.

Step 3: Determine Form for (a)

Since neither linear nor exponential characteristics are observed, the table in (a) does not fit a linear or exponential function.

Step 4: Examine Linear Function for (b)

Check if (b) is a linear function by comparing differences between consecutive y-values:- Between -1 and 0: \(30.12 - 50.2 = -20.08\)- Between 0 and 1: \(18.072 - 30.12 = -12.048\)- Between 1 and 2: \(10.8432 - 18.072 = -7.2288\)Differences are not constant, so (b) is not a linear function.

Step 5: Examine Exponential Function for (b)

Compute the ratio of consecutive values to check for exponential characteristics:- Between -1 and 0: \(\frac{30.12}{50.2} = 0.6\)- Between 0 and 1: \(\frac{18.072}{30.12} = 0.6\)- Between 1 and 2: \(\frac{10.8432}{18.072} = 0.6\)Since the ratio is constant, (b) is an exponential function.

Step 6: Determine Exponential Formula for (b)

Using the exponential form \(s(t) = a \, b^t\), where \(s(-1) = 50.2\), we know:- \(b = 0.6\) (common ratio)- Calculate \(a\) using initial value: \(s(0) = a \cdot 0.6^{0} = 30.12\), so \(a = 30.12\).The exponential formula is: \[s(t) = 50.2 \times 0.6^{t+1}\]

Step 7: Examine Linear Function for (c)

Check if (c) is linear by calculating differences:- Between 0 and 2: \(24 - 27 = -3\)- Between 2 and 4: \(21 - 24 = -3\)- Between 4 and 6: \(18 - 21 = -3\)Since the differences are constant, (c) follows a linear pattern.

Step 8: Determine Linear Formula for (c)

Using the form \(g(u) = mu + c\), where \(m=-1.5\) (as shown by the difference \(-3\) over span of 2 -units).- We calculate the intercept \(c\) using one point, e.g., \(g(0) = 27\), leading to \(c = 27\).Thus, the linear function is:\[g(u) = -1.5u + 27\]

Key Concepts

Linear Function

In mathematics, a linear function is a simple function that creates a straight line when graphed. It is expressed in the form of \( f(x) = mx + c \), where \( m \) represents the slope and \( c \) is the y-intercept. Understanding linear functions is important because they can model real-world situations like calculating earnings or basic budget planning.
To determine if data represents a linear function, check the differences between consecutive values. If these differences are consistent, then the data is likely modeling a linear function.
For example, in table (c) of the exercise, the differences are \(-3\) consistently between each pair of points. Thus, it fits the form of a linear function, with a rate of change \( m = -1.5 \). Calculating the intercept from a known point, such as \( g(0) = 27 \), allows us to completely define the linear equation as \( g(u) = -1.5u + 27 \).

• Linear functions show constant addition or subtraction between values, resulting in a straight line on a graph.
• Recognizing linear equations helps in modeling scenarios with consistent change or rates.

Exponential Function

Exponential functions are distinct for their multiplicative growth or decay, characterized by a constant ratio between successive values. They have the general form \( f(x) = a \, b^x \), where \( a \) is the initial value and \( b \) is the base or growth factor. If \( b > 1 \), the function represents growth, while a value between zero and one shows decay.
In problem (b) from the exercise, we determine that this table represents an exponential function because the ratio between consecutive values is constant at 0.6. This indicates a decay pattern, confirming its exponential nature.
To find the function formula, we identify \( a \) and \( b \), where \( a = 50.2 \), derived from the context, and confirmed by the initial entry, and \( b = 0.6 \), as calculated from the consistent ratio. The resulting exponential equation is \( s(t) = 50.2 \times 0.6^{t+1} \).

• Exponential functions model patterns where values change by a consistent percentage or factor.
• Such functions are crucial for understanding processes involving growth or decay, like population dynamics or radioactive decay.

Table of Values

Tables of values are powerful tools used to analyze and interpret the behaviors of various functions. They provide a clear, structured presentation of the relationship between variables, making it easier to see patterns or predict trends.
Choosing whether a table corresponds to a linear or exponential function requires examining the differences or ratios between values. For linear functions, as seen in example (c), look for consistent additions or subtractions between successive entries. Meanwhile, exponential functions, like in (b), show consistency in the ratios of consecutive terms.
Using tables to define functions involves recognizing these patterns, which then aids in creating models that describe relationships between variables in a formulaic manner.

• Tables help identify function types by revealing patterns in data sequences.
• A logical approach to analyzing tables enhances prediction of future values based on recognized trends.