Question

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)

Accepted Answer

Identify the behavior of each function as $$ x 	o \infty . $$For $$ f(x) = 3^{-x} $$, rewrite it as $$ f(x) = \left( \frac{1}{3} ight)^x . $$Similarly, for $$ g(x) = 2^{-x} $$, rewrite it as $$ g(x) = \left( \frac{1}{2} ight)^x . $$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 $$ x 	o \infty . To $$compare their rates of decay, consider the bases: $$ \frac{1}{3} \approx 0.333 $$ and $$ \frac{1}{2} = 0.5 . $$Since $$ 0.333 \u003c 0.5 $$, $$ f(x) $$ decreases faster than $$ g(x) . $$Another way to compare growth rates is to look at the limit of their ratio as $$ x 	o \infty $$: calculate $$ \lim_{x 	o \infty} \frac{f(x)}{g(x)} = \lim_{x 	o \infty} \frac{(1/3)^x}{(1/2)^x} = \lim_{x 	o \infty} \left( \frac{1/3}{1/2} ight)^x = \lim_{x 	o \infty} \left( \frac{2}{3} ight)^x . $$Since $$ \frac{2}{3} \u003c 1 $$, this limit approaches 0, which confirms that $$ f(x) $$ goes to zero faster than $$ g(x)  as$$ $$ x 	o \infty . $$Therefore, $$ f $$ grows slower (decays faster) than $$ g  as$$ $$ x 	o \infty .$$

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)

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Exponential Functions and Their Growth Rates

Limits at Infinity

Comparing Rates of Decay Using Dominance