The function you've provided is \( f(x) = 40,000(0.85)^x \).
This function represents an exponential decay model where:
- The initial value (when \( x = 0 \)) is 40,000.
- The base of the exponent (0.85) suggests that the value decreases by 15% each time \( x \) increases by 1.
Key Points:
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Initial Value: When \( x = 0 \): \[ f(0) = 40,000(0.85)^0 = 40,000 \cdot 1 = 40,000 \]
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Decay Rate: The function decreases because the base (0.85) is less than 1. For each unit increase in \( x \), the function's value is multiplied by 0.85.
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Long-Term Behavior: As \( x \) increases, \( f(x) \) approaches 0 but never actually reaches it due to the nature of exponential functions.
Example Calculations:
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When \( x = 1 \): \[ f(1) = 40,000(0.85)^1 = 40,000 \cdot 0.85 = 34,000 \]
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When \( x = 2 \): \[ f(2) = 40,000(0.85)^2 = 40,000 \cdot 0.7225 = 28,900 \]
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When \( x = 3 \): \[ f(3) = 40,000(0.85)^3 = 40,000 \cdot 0.614125 = 24,565 \]
This function can be used for various applications, such as modeling population decline, depreciation of an asset, or similar scenarios where quantities decrease over time. If you have any specific questions or further details you’d like to explore, please let me know!