Question

f(x)= 64(0.50)^x
At a tournament teams play each other with only the winning team progressing to the next round. In other words the number of teams playing at each round is half of the number of teams playing in the previous round. Define an appropriate range for this situation

Answer 1

In the given function \( f(x) = 64(0.50)^x \), \( f(x) \) represents the number of teams remaining after \( x \) rounds of a tournament where teams are eliminated in pairs (i.e., only the winning teams progress to the next round).

To define an appropriate range for the situation:

1. **Initial Teams**: At the start of the tournament (when \( x = 0 \)), \( f(0) = 64(0.50)^0 = 64 \) teams play.

2. **Number of Teams**: As each round progresses, the number of teams halves:
- After 1 round (\( x = 1 \)): \( f(1) = 64(0.50)^1 = 32 \)
- After 2 rounds (\( x = 2 \)): \( f(2) = 64(0.50)^2 = 16 \)
- After 3 rounds (\( x = 3 \)): \( f(3) = 64(0.50)^3 = 8 \)
- After 4 rounds (\( x = 4 \)): \( f(4) = 64(0.50)^4 = 4 \)
- After 5 rounds (\( x = 5 \)): \( f(5) = 64(0.50)^5 = 2 \)
- After 6 rounds (\( x = 6 \)): \( f(6) = 64(0.50)^6 = 1 \)
- After 7 rounds (\( x = 7 \)): \( f(7) = 64(0.50)^7 = 0.5 \)

3. **Termination of the Tournament**: The tournament logically ends when there is at least 1 team left. We cannot have a fraction of a team, so \( f(x) \) must be an integer.

Thus, from the calculations above, we see that:
- The maximum number of teams is 64 (at \( x = 0 \)).
- The minimum number of teams that can be considered to still be participating in the tournament is 1 (at \( x = 6 \)).
- Beyond \( x = 6 \), \( f(x) \) produces fractional values, indicating that a team cannot be halved further in the context of this tournament setup.

Therefore, the appropriate range for \( f(x) \) is the set of integers from 1 to 64.

In interval notation, this can be written as:
\[ \text{Range} = \{ 64, 32, 16, 8, 4, 2, 1 \} \] or in terms of integers:
\[ \text{Range} = \{ n \in \mathbb{Z} \mid 1 \leq n \leq 64 \} \]