Find the domain and range of the function k(x) = (x+1) / sqrt(2x+1)

Step 1: Find the domain The function is defined when the expression inside the square root is positive and the denominator is not zero: 2x + 1 > 0 Solve the inequality: 2x > -1 x > -(1)/(2) Domain: ( -(1)/(2), ) Step 2: Find the range Let y = k(x) = (x+1)/(sqrt(2x+1)) . Substitute t = sqrt(2x+1) where t > 0 : 2x + 1 = t^2 x = (t^2 - 1)/(2) Substitute into y : y = (t^2 - 1)/(2) + 1t = (t^2 - 1 + 2)/(2)t = (t^2 + 1)/(2)t = (t^2 + 1)/(2t) y = (t)/(2) + (1)/(2t), t > 0 To find the range, analyze f(t) = (t)/(2) + (1)/(2t) for t > 0 . Find the minimum using calculus or AM-GM inequality: By AM-GM: (t)/(2) + (1)/(2t) ≥ 2sqrt((t)/(2) · (1)/(2t)) = 2sqrt((1)/(4)) = 1 Equality when (t)/(2) = (1)/(2t) , so t^2 = 1 , t = 1 (since t > 0 ). At t = 1 , y = 1 . As t 0^+ , y . As t , y . Thus, y ≥ 1 . Range: [1, ) Final answer c Domain: ( -(1)/(2),\ ) \\[2em] Range: [1,\ )