Question 15 Find the range and domain of the function \( f(x)=4\left(\frac{1}{2}\right)^{x}+5 \) Domain: \( ? \in \in \) ? Range: \( ? \in \in \square \)

To find the domain and range of the function \( f(x) = 4\left(\frac{1}{2}\right)^{x} + 5 \), we can analyze the function step by step. ### Domain The function \( f(x) \) is defined for all real numbers \( x \) because there are no restrictions (like division by zero or square roots of negative numbers). Therefore, the domain is: \[ \text{Domain: } x \in (-\infty, \infty) \] ### Range Next, we analyze the range of the function. The term \( \left(\frac{1}{2}\right)^{x} \) is an exponential function. As \( x \) approaches \( -\infty \), \( \left(\frac{1}{2}\right)^{x} \) approaches \( +\infty \). As \( x \) approaches \( +\infty \), \( \left(\frac{1}{2}\right)^{x} \) approaches \( 0 \). Thus, the function \( f(x) \) behaves as follows: - As \( x \to -\infty \), \( f(x) \to 4 \cdot (+\infty) + 5 = +\infty \). - As \( x \to +\infty \), \( f(x) \to 4 \cdot 0 + 5 = 5 \). Since \( f(x) \) approaches \( 5 \) but never actually reaches it (as \( \left(\frac{1}{2}\right)^{x} \) is always positive), the minimum value of \( f(x) \) is \( 5 \), and it can take any value greater than \( 5 \). Therefore, the range of the function is: \[ \text{Range: } f(x) \in (5, +\infty) \] ### Summary - **Domain:** \( x \in (-\infty, \infty) \) - **Range:** \( f(x) \in (5, +\infty) \)