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What is the domain of the given exponential function below? \( 2\left(\frac{1}{2}\right)^{n+1}+2 \) \( (2, \infty) \) \( [2, \infty) \) \( (-\infty, \infty) \) \( (1, \infty) \)

Ask by Pollard Ward. in the United States
Jan 25,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The domain of the function \( 2\left(\frac{1}{2}\right)^{n+1}+2 \) is all real numbers, so the correct choice is \( (-\infty, \infty) \).

Solution

To determine the domain of the given exponential function \( f(n) = 2\left(\frac{1}{2}\right)^{n+1}+2 \), we need to analyze the function. 1. **Identify the function type**: The function is an exponential function, which is defined for all real numbers. The base \( \frac{1}{2} \) is a positive number, and the exponent \( n+1 \) can take any real value. 2. **Exponential function properties**: Exponential functions of the form \( a^x \) (where \( a > 0 \)) are defined for all real numbers \( x \). Therefore, \( \left(\frac{1}{2}\right)^{n+1} \) is defined for all \( n \). 3. **Adding constants**: The function also includes a constant term \( +2 \), which does not affect the domain. Since there are no restrictions on \( n \) in the expression, the domain of the function is all real numbers. Thus, the domain of the function \( 2\left(\frac{1}{2}\right)^{n+1}+2 \) is: \[ (-\infty, \infty) \] The correct answer is \( (-\infty, \infty) \).

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The domain of the function \( 2\left(\frac{1}{2}\right)^{n+1}+2 \) is all real numbers, as there are no restrictions on the variable \( n \) in the exponent. Exponential functions can take any real number as input and will produce a valid output for all of them. So, the correct answer is \( (-\infty, \infty) \). This means that you can input any value, and the function will always give you a corresponding output! Isn’t that neat? There’s a lot of flexibility when you’re dealing with exponentials! In practical applications, these kinds of functions are often used in scenarios like population growth, radioactive decay, or even in finance when calculating compound interest. No matter what value of \( n \) you choose, the function models a consistent pattern that grows or shrinks exponentially, which can be quite fascinating to explore!

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