Find the sum function \( (f+g)(x) \) if \( f(x)=\left\{\begin{array}{ll}2 x+5 & \text { if } x<1 \\ x^{2}+3 x & \text { if } x \geq 1\end{array}\right. \) and \( g(x)=\left\{\begin{array}{rr}-4 x+1 & \text { if } x \leq 0 \\ x-6 & \text { if } x>0\end{array}\right. \) Select the correct choice below and fill in the answer boxes to complete your choice. A. \( (f+g)(x)=\left\{\begin{array}{ll}\square & \text { if } x \leq \square \\ \square & \text { if } x>\square\end{array}\right. \) (Simplify your answers. Do not factor.) B. \( (f+g)(x)=\left\{\begin{array}{ll}\square & \text { if } x \leq \square \\ \square & \text { if } \square

Ask by Schmidt Hobbs. in the United States

Feb 26,2025

We begin with the definitions:   f(x) = { 2x + 5 if x < 1,        x² + 3x if x ≥ 1 }   g(x) = { –4x + 1 if x ≤ 0,        x – 6 if x > 0 } Notice that the pieces change at x = 0 (for g) and x = 1 (for f). Therefore, we must break the real line into three regions: 1. When x ≤ 0 (here, f uses its x < 1 definition and g uses its x ≤ 0 definition). 2. When 0 < x < 1 (f still uses its x < 1 definition, but now g uses its x > 0 definition). 3. When x ≥ 1 (now f changes to its x ≥ 1 part and g continues to use its x > 0 piece since 1 > 0). Now, compute the sum (f + g)(x) in each region. Region 1: For x ≤ 0,   f(x) = 2x + 5      (given that x < 1)   g(x) = –4x + 1 So,   (f + g)(x) = (2x + 5) + (–4x + 1) = –2x + 6 Region 2: For 0 < x < 1,   f(x) = 2x + 5       (again, we use the piece for x < 1)   g(x) = x – 6 Thus,   (f + g)(x) = (2x + 5) + (x – 6) = 3x – 1 Region 3: For x ≥ 1,   f(x) = x² + 3x       (use the definition for x ≥ 1)   g(x) = x – 6        (since x > 0) So,   (f + g)(x) = (x² + 3x) + (x – 6) = x² + 4x – 6 Thus the sum function is given by:   (f + g)(x) = { –2x + 6  if x ≤ 0,            3x – 1  if 0 < x < 1,            x² + 4x – 6 if x ≥ 1 } This corresponds to answer choice B with the answer boxes filled as follows: B. (f + g)(x) = { –2x + 6 if x ≤ 0,       3x – 1 if 0 < x < 1,       x² + 4x – 6 if x ≥ 1 }