It is given that \[ f(x)=\left\{\begin{array}{ll}7-x^{2} & \text { for } 0

Ask by Rodriguez Boyd. in Singapore

Jan 22,2025

To evaluate \( f(27) + f(45) \), we need to utilize the periodicity of the function \( f(x) = f(x + 4) \). This means that for any \( x \), we can reduce \( x \) modulo 4 to find its equivalent value for evaluating \( f \). We begin with \( f(27) \): \[ 27 \mod 4 = 3 \quad (\text{since } 27 = 4 \times 6 + 3) \] Thus, \( f(27) = f(3) \). Next, we find \( f(3) \) using the piecewise definition: Since \( 2 < 3 \leq 4 \), we use the second case: \[ f(3) = 2 \cdot 3 - 1 = 6 - 1 = 5. \] Now, we compute \( f(45) \): \[ 45 \mod 4 = 1 \quad (\text{since } 45 = 4 \times 11 + 1) \] Thus, \( f(45) = f(1) \). Now we find \( f(1) \) using the piecewise definition: Since \( 0 < 1 \leq 2 \), we use the first case: \[ f(1) = 7 - 1^2 = 7 - 1 = 6. \] Finally, we add the results: \[ f(27) + f(45) = f(3) + f(1) = 5 + 6 = 11. \] Thus, the answer is: \[ \boxed{11}. \]