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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.PE.110a
Chapter 7, Problem 7.PE.110a

110. Does f grow faster, slower, or at the same rate as g as x→∞? Give reasons for your answers.
a. f(x) = 3^(-x), g(x) = 2^(-x)

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Identify the behavior of each function as \( x \to \infty \). For \( f(x) = 3^{-x} \), rewrite it as \( f(x) = \left( \frac{1}{3} \right)^x \). Similarly, for \( g(x) = 2^{-x} \), rewrite it as \( g(x) = \left( \frac{1}{2} \right)^x \).
Recognize that both functions are exponential decay functions because their bases \( \frac{1}{3} \) and \( \frac{1}{2} \) are between 0 and 1, so both approach 0 as \( x \to \infty \).
To compare their rates of decay, consider the bases: \( \frac{1}{3} \approx 0.333 \) and \( \frac{1}{2} = 0.5 \). Since \( 0.333 < 0.5 \), \( f(x) \) decreases faster than \( g(x) \).
Another way to compare growth rates is to look at the limit of their ratio as \( x \to \infty \): calculate \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{(1/3)^x}{(1/2)^x} = \lim_{x \to \infty} \left( \frac{1/3}{1/2} \right)^x = \lim_{x \to \infty} \left( \frac{2}{3} \right)^x \).
Since \( \frac{2}{3} < 1 \), this limit approaches 0, which confirms that \( f(x) \) goes to zero faster than \( g(x) \) as \( x \to \infty \). Therefore, \( f \) grows slower (decays faster) than \( g \) as \( x \to \infty \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential Functions and Their Growth Rates

Exponential functions have the form a^x, where the base a determines the growth or decay rate. For bases between 0 and 1, the function represents exponential decay, decreasing as x increases. Comparing two such functions involves analyzing their bases to determine which decays faster.
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Limits at Infinity

The limit of a function as x approaches infinity describes its long-term behavior. For exponential decay functions, the limit approaches zero, but the rate at which they approach zero depends on the base. Understanding limits helps compare how quickly functions approach zero.
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Comparing Rates of Decay Using Dominance

When comparing two functions as x→∞, the function with the smaller base in exponential decay decreases faster. This concept of dominance helps determine which function grows slower or faster by examining which tends to zero more rapidly.
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