Answer :
Let's examine both parts of the problem step-by-step:
### Part A: Analyzing the Price of Fuel A
The price \( f(x) \) of fuel A after \( x \) months is given by the function:
[tex]\[ f(x) = 2.15(0.98)^x \][/tex]
1. Determine if the price is increasing or decreasing:
- The function \( f(x) = 2.15(0.98)^x \) is in the form of an exponential decay function because the base of the exponent (0.98) is less than 1.
- This indicates that the price of fuel A is decreasing over time.
2. Find the rate of percentage decrease per month:
- In exponential decay functions of the form \( f(x) = a \cdot b^x \), where \( 0 < b < 1 \), the base \( b \) represents \( 1 - r \), where \( r \) is the rate of decrease.
- Here, \( b = 0.98 \), hence \( 1 - r = 0.98 \).
- Solving for \( r \), we get \( r = 1 - 0.98 = 0.02 \).
- Converting this to a percentage gives us a 2% decrease per month.
Therefore, the price of fuel A is decreasing by 2% per month.
### Part B: Analyzing the Price of Fuel B
The table below shows the price \( g(m) \) of fuel B after \( m \) months:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline m \text{ (number of months)} & 1 & 2 & 3 & 4 \\ \hline g(m) \text{ (price in dollars)} & 4.19 & 3.98 & 3.78 & 3.59 \\ \hline \end{array} \][/tex]
1. Calculate the percentage change in price for fuel B from month-to-month:
The percentage change from month \( m \) to month \( m+1 \) is calculated as:
[tex]\[ \text{Percentage change} = \left( \frac{g(m) - g(m+1)}{g(m)} \right) \times 100 \][/tex]
- From month 1 to month 2:
[tex]\[ \left( \frac{4.19 - 3.98}{4.19} \right) \times 100 = 5.0119\% \][/tex]
- From month 2 to month 3:
[tex]\[ \left( \frac{3.98 - 3.78}{3.98} \right) \times 100 = 5.0251\% \][/tex]
- From month 3 to month 4:
[tex]\[ \left( \frac{3.78 - 3.59}{3.78} \right) \times 100 = 5.0265\% \][/tex]
The percentage changes are approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \).
2. Find the greatest percentage change:
- Among the calculated percentage changes, the greatest percentage change is \( 5.0265\% \), which occurred from month 3 to month 4.
### Comparisons and Conclusion:
- Fuel A: The price is decreasing steadily by \( 2\% \) per month.
- Fuel B: The price has recorded percentage changes of approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \) over the previous months.
Conclusion: Fuel B has recorded a greater percentage change in price over the previous month compared to Fuel A. The maximum percentage change for Fuel B recorded is approximately 5.0265%, which is greater than the steady 2% decrease per month for Fuel A.
### Part A: Analyzing the Price of Fuel A
The price \( f(x) \) of fuel A after \( x \) months is given by the function:
[tex]\[ f(x) = 2.15(0.98)^x \][/tex]
1. Determine if the price is increasing or decreasing:
- The function \( f(x) = 2.15(0.98)^x \) is in the form of an exponential decay function because the base of the exponent (0.98) is less than 1.
- This indicates that the price of fuel A is decreasing over time.
2. Find the rate of percentage decrease per month:
- In exponential decay functions of the form \( f(x) = a \cdot b^x \), where \( 0 < b < 1 \), the base \( b \) represents \( 1 - r \), where \( r \) is the rate of decrease.
- Here, \( b = 0.98 \), hence \( 1 - r = 0.98 \).
- Solving for \( r \), we get \( r = 1 - 0.98 = 0.02 \).
- Converting this to a percentage gives us a 2% decrease per month.
Therefore, the price of fuel A is decreasing by 2% per month.
### Part B: Analyzing the Price of Fuel B
The table below shows the price \( g(m) \) of fuel B after \( m \) months:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline m \text{ (number of months)} & 1 & 2 & 3 & 4 \\ \hline g(m) \text{ (price in dollars)} & 4.19 & 3.98 & 3.78 & 3.59 \\ \hline \end{array} \][/tex]
1. Calculate the percentage change in price for fuel B from month-to-month:
The percentage change from month \( m \) to month \( m+1 \) is calculated as:
[tex]\[ \text{Percentage change} = \left( \frac{g(m) - g(m+1)}{g(m)} \right) \times 100 \][/tex]
- From month 1 to month 2:
[tex]\[ \left( \frac{4.19 - 3.98}{4.19} \right) \times 100 = 5.0119\% \][/tex]
- From month 2 to month 3:
[tex]\[ \left( \frac{3.98 - 3.78}{3.98} \right) \times 100 = 5.0251\% \][/tex]
- From month 3 to month 4:
[tex]\[ \left( \frac{3.78 - 3.59}{3.78} \right) \times 100 = 5.0265\% \][/tex]
The percentage changes are approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \).
2. Find the greatest percentage change:
- Among the calculated percentage changes, the greatest percentage change is \( 5.0265\% \), which occurred from month 3 to month 4.
### Comparisons and Conclusion:
- Fuel A: The price is decreasing steadily by \( 2\% \) per month.
- Fuel B: The price has recorded percentage changes of approximately \( 5.0119\% \), \( 5.0251\% \), and \( 5.0265\% \) over the previous months.
Conclusion: Fuel B has recorded a greater percentage change in price over the previous month compared to Fuel A. The maximum percentage change for Fuel B recorded is approximately 5.0265%, which is greater than the steady 2% decrease per month for Fuel A.
