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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 121b

Exponential growth rates
b. Compare the growth rates of eˣ and eᵃˣ as x→∞ , for a > 0.

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To compare the growth rates of \( e^x \) and \( e^{ax} \) as \( x \to \infty \), we first need to understand the behavior of exponential functions. Both functions are exponential, but they have different exponents.
Consider the function \( e^x \). As \( x \to \infty \), \( e^x \) grows exponentially without bound. The base \( e \) is a constant greater than 1, which means the function increases rapidly.
Now, consider the function \( e^{ax} \). Here, \( a \) is a positive constant greater than 0. The exponent \( ax \) means that the rate of growth is scaled by \( a \). If \( a > 1 \), \( e^{ax} \) grows faster than \( e^x \). If \( 0 < a < 1 \), \( e^{ax} \) grows slower than \( e^x \).
To compare the growth rates more formally, we can take the limit of the ratio of the two functions as \( x \to \infty \): \( \lim_{x \to \infty} \frac{e^{ax}}{e^x} = \lim_{x \to \infty} e^{(a-1)x} \).
Evaluate the limit: If \( a > 1 \), \( e^{(a-1)x} \to \infty \), indicating \( e^{ax} \) grows faster. If \( 0 < a < 1 \), \( e^{(a-1)x} \to 0 \), indicating \( e^x \) grows faster. If \( a = 1 \), the limit is 1, indicating both grow at the same rate.

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