Asked by Tesla
In exponential growth functions, the base of the exponent must be greater than 1. How would the function change if the base of the exponent were 1? How would the function change if the base of the exponent were between 0 and 1?
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GPT-5 mini
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Take a general exponential function y = a·b^x with b>0.
- If b = 1: b^x = 1 for every x, so y = a·1 = a. The exponential becomes a constant function (no growth or decay), graph is a horizontal line y = a.
- If 0 < b < 1: the function is exponential decay. Each increase of x by 1 multiplies y by b (<1), so y decreases as x increases. For a > 0:
- y → 0 as x → +∞ (horizontal asymptote y = 0),
- y → +∞ as x → −∞.
Example: y = (1/2)^x = 2^(−x) is a decreasing curve. Equivalently, b^x with 0<b<1 is the same shape as (1/b)^{−x} (a growth function reflected across the y-axis).
(As a side note: negative bases are not used for real-valued exponentials with arbitrary real x because they are generally not real-valued; b must be positive if x ranges over the reals.)
- If b = 1: b^x = 1 for every x, so y = a·1 = a. The exponential becomes a constant function (no growth or decay), graph is a horizontal line y = a.
- If 0 < b < 1: the function is exponential decay. Each increase of x by 1 multiplies y by b (<1), so y decreases as x increases. For a > 0:
- y → 0 as x → +∞ (horizontal asymptote y = 0),
- y → +∞ as x → −∞.
Example: y = (1/2)^x = 2^(−x) is a decreasing curve. Equivalently, b^x with 0<b<1 is the same shape as (1/b)^{−x} (a growth function reflected across the y-axis).
(As a side note: negative bases are not used for real-valued exponentials with arbitrary real x because they are generally not real-valued; b must be positive if x ranges over the reals.)
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