Answer :
Certainly! Let's break down and interpret the given function step-by-step:
### Step 1: Understanding the Initial Population
The given function is [tex]\( f(x) = 50,000 \cdot (0.98)^x \)[/tex], where [tex]\( x \)[/tex] represents the number of years.
- The initial population of Frillsville is 50,000. This is the starting point, meaning when [tex]\( x = 0 \)[/tex], we have [tex]\( f(0) = 50,000 \cdot (0.98)^0 \)[/tex]. Since any number raised to the power of 0 is 1, it simplifies to [tex]\( f(0) = 50,000 \)[/tex].
### Step 2: Understanding the Change Factor
- The change factor is 0.98. This means that each year, the population is multiplied by 0.98.
### Step 3: Yearly Population Change
To understand the change, consider what multiplying by 0.98 implies:
- If we take 0.98 as a percentage, it translates to 98%. Hence, the population retains 98% of its value from the previous year.
- Conversely, this means there is a 2% decline each year.
### Step 4: Predicting the Population Over Time
For any given year [tex]\( x \)[/tex]:
- The population decreases exponentially according to the formula [tex]\( f(x) = 50,000 \cdot (0.98)^x \)[/tex].
For example:
- After 1 year ([tex]\( x=1 \)[/tex]): the population is [tex]\( f(1) = 50,000 \cdot 0.98 \approx 49,000 \)[/tex].
- After 2 years ([tex]\( x=2 \)[/tex]): the population is [tex]\( f(2) = 50,000 \cdot (0.98)^2 \approx 48,020 \)[/tex].
### Step 5: Long-Term Trends
As [tex]\( x \)[/tex] increases:
- The term [tex]\( (0.98)^x \)[/tex] gets smaller because raising a number less than 1 to increasing power reduces its value.
- This indicates a consistent annual decrease in population, validating the 2% reduction per year.
### Conclusion
- The population model [tex]\( f(x) = 50,000 \cdot (0.98)^x \)[/tex] tells us that the population of Frillsville starts at 50,000 and decreases by 2% each year.
- Over time, the population continues to decline, following this exponential decay pattern.
### Step 1: Understanding the Initial Population
The given function is [tex]\( f(x) = 50,000 \cdot (0.98)^x \)[/tex], where [tex]\( x \)[/tex] represents the number of years.
- The initial population of Frillsville is 50,000. This is the starting point, meaning when [tex]\( x = 0 \)[/tex], we have [tex]\( f(0) = 50,000 \cdot (0.98)^0 \)[/tex]. Since any number raised to the power of 0 is 1, it simplifies to [tex]\( f(0) = 50,000 \)[/tex].
### Step 2: Understanding the Change Factor
- The change factor is 0.98. This means that each year, the population is multiplied by 0.98.
### Step 3: Yearly Population Change
To understand the change, consider what multiplying by 0.98 implies:
- If we take 0.98 as a percentage, it translates to 98%. Hence, the population retains 98% of its value from the previous year.
- Conversely, this means there is a 2% decline each year.
### Step 4: Predicting the Population Over Time
For any given year [tex]\( x \)[/tex]:
- The population decreases exponentially according to the formula [tex]\( f(x) = 50,000 \cdot (0.98)^x \)[/tex].
For example:
- After 1 year ([tex]\( x=1 \)[/tex]): the population is [tex]\( f(1) = 50,000 \cdot 0.98 \approx 49,000 \)[/tex].
- After 2 years ([tex]\( x=2 \)[/tex]): the population is [tex]\( f(2) = 50,000 \cdot (0.98)^2 \approx 48,020 \)[/tex].
### Step 5: Long-Term Trends
As [tex]\( x \)[/tex] increases:
- The term [tex]\( (0.98)^x \)[/tex] gets smaller because raising a number less than 1 to increasing power reduces its value.
- This indicates a consistent annual decrease in population, validating the 2% reduction per year.
### Conclusion
- The population model [tex]\( f(x) = 50,000 \cdot (0.98)^x \)[/tex] tells us that the population of Frillsville starts at 50,000 and decreases by 2% each year.
- Over time, the population continues to decline, following this exponential decay pattern.
Step 1: Understanding the Initial Population
The given function is \( f(x) = 50,000 \cdot (0.98)^x \), where \( x \) represents the number of years.
- The initial population of Frillsville is 50,000. This is the starting point, meaning when \( x = 0 \), we have \( f(0) = 50,000 \cdot (0.98)^0 \). Since any number raised to the power of 0 is 1, it simplifies to \( f(0) = 50,000 \).
Step 2: Understanding the Change Factor
- The change factor is 0.98. This means that each year, the population is multiplied by 0.98.
Step 3: Yearly Population Change
To understand the change, consider what multiplying by 0.98 implies:
- If we take 0.98 as a percentage, it translates to 98%. Hence, the population retains 98% of its value from the previous year.
- Conversely, this means there is a 2% decline each year.
Step 4: Predicting the Population Over Time
For any given year \( x \):
- The population decreases exponentially according to the formula \( f(x) = 50,000 \cdot (0.98)^x \).
For example:
- After 1 year (\( x=1 \)): the population is \( f(1) = 50,000 \cdot 0.98 \approx 49,000 \).
- After 2 years (\( x=2 \)): the population is \( f(2) = 50,000 \cdot (0.98)^2 \approx 48,020 \).
Step 5: Long-Term Trends
As \( x \) increases:
- The term \( (0.98)^x \) gets smaller because raising a number less than 1 to increasing power reduces its value.
- This indicates a consistent annual decrease in population, validating the 2% reduction per year.
Answer:
- The population model \( f(x) = 50,000 \cdot (0.98)^x \) tells us that the population of Frillsville starts at 50,000 and decreases by 2% each year.
- Over time, the population continues to decline, following this exponential decay pattern.
