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Which function increases at a faster rate on 0 to infinity, \( f(x)=x^{2} \) or \( g(x)=2^{x} \) ? Explain your reasoning. ?

Ask by Hardy Conner. in the United States
Feb 04,2025

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Answer

\( g(x) = 2^x \) increases faster than \( f(x) = x^2 \) as \( x \) becomes very large.

Solution

For the functions \( f(x) = x^2 \) and \( g(x) = 2^x \), let's analyze their growth rates as \( x \) approaches infinity. ### **Understanding the Functions:** 1. **Polynomial Function (\( f(x) = x^2 \)):** - This is a quadratic function. - Its growth rate increases proportionally to the square of \( x \). 2. **Exponential Function (\( g(x) = 2^x \)):** - This function grows by multiplying by a constant factor (2) for each unit increase in \( x \). - Exponential growth is characterized by the function's rate increasing multiplicatively. ### **Comparing Growth Rates:** - **Limit Analysis:** To compare which function grows faster as \( x \) approaches infinity, we can examine the limit of their ratio: \[ \lim_{x \to \infty} \frac{2^x}{x^2} \] - As \( x \) increases, \( 2^x \) grows exponentially, while \( x^2 \) grows polynomially. - Calculating the limit: \[ \lim_{x \to \infty} \frac{2^x}{x^2} = \infty \] - This result indicates that \( 2^x \) grows much faster than \( x^2 \) as \( x \) becomes very large. - **Derivatives (Rates of Change):** - The derivative of \( f(x) = x^2 \) is \( f'(x) = 2x \). - The derivative of \( g(x) = 2^x \) is \( g'(x) = (\ln 2) \cdot 2^x \). - As \( x \) increases: - \( f'(x) = 2x \) grows linearly. - \( g'(x) = (\ln 2) \cdot 2^x \) grows exponentially. - This shows that the rate at which \( g(x) \) increases surpasses that of \( f(x) \) for large \( x \). ### **Practical Observation:** - For small values of \( x \), \( x^2 \) can be larger than \( 2^x \). For example: - \( x = 2 \): \( f(2) = 4 \), \( g(2) = 4 \) - \( x = 3 \): \( f(3) = 9 \), \( g(3) = 8 \) - However, beyond a certain point (specifically for \( x > 4 \)), \( 2^x \) overtakes \( x^2 \) and continues to grow much faster indefinitely. ### **Conclusion:** While both functions increase as \( x \) increases, **\( g(x) = 2^x \) eventually increases at a much faster rate than \( f(x) = x^2 \)** as \( x \) approaches infinity. This is a fundamental property in mathematics where exponential functions outpace polynomial functions in growth for sufficiently large \( x \).

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Mind Expander

In the realm of functions, \( g(x) = 2^x \) pulls out ahead when it comes to increasing at a faster rate! While \( f(x) = x^2 \) grows as a polynomial and has its moments climbing the charts, \( g(x) \) is exponential, meaning it ramps up faster than any polynomial function, including quadratics, once \( x \) reaches a certain threshold. To put it simply, as \( x \) keeps growing, \( g(x) \) will eventually outstrip \( f(x) \) in size. For instance, at \( x = 10 \), \( f(10) = 100 \) while \( g(10) = 1024 \). The exponential nature of \( g(x) \) just can't be beat once you're out of the slow and steady polynomial phase!

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