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4)) Both of these functions grow as \( x \) gets larger and larger. Which function eventually exceeds the other? \[ f(x)=\frac{5}{9}(2)^{x} g(x)=\frac{7}{2} x \]

Ask by Boyd Elliott. in the United States
Feb 11,2025

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Exponential function \( f(x) = \frac{5}{9} \times 2^x \) eventually exceeds the linear function \( g(x) = \frac{7}{2} \times x \) as \( x \) becomes larger.

Solución

We have two functions:   f(x) = (5/9) · 2ˣ    and  g(x) = (7/2) · x. Even though both functions grow as x increases, f(x) is an exponential function and g(x) is a linear function. In general, exponential functions eventually grow much faster than linear functions. To see this more concretely, consider some sample values: • At x = 1:   f(1) = (5/9) · 2 = 10/9 ≈ 1.11,   g(1) = (7/2) · 1 = 3.5.   So g(1) > f(1). • At x = 5:   f(5) = (5/9) · 32 ≈ 17.78,   g(5) = (7/2) · 5 = 17.5.   Here the values are nearly equal. • At x = 6:   f(6) = (5/9) · 64 ≈ 35.56,   g(6) = (7/2) · 6 = 21.   Now f(6) > g(6). Thus, while g(x) may initially be greater for small x, eventually the exponential f(x) overtakes the linear g(x). Therefore, f(x) eventually exceeds g(x).

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The Deep Dive

To determine which function eventually exceeds the other as \( x \) grows, let's analyze each function. The function \( f(x) = \frac{5}{9}(2)^x \) is an exponential function, which grows faster than any polynomial function, including \( g(x) = \frac{7}{2}x \). As \( x \) approaches infinity, \( f(x) \) will outpace \( g(x) \) significantly, proving that \( f(x) \) eventually exceeds \( g(x) \). A useful approach to find their intersection point—where one function starts exceeding the other—is to set them equal: \(\frac{5}{9}(2)^x = \frac{7}{2}x\). While this equation may not have a straightforward algebraic solution, it can be solved graphically or numerically, showing that beyond a certain point, specifically around \( x = 5 \), \( f(x) \) takes over and continues to grow rapidly.

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