The following function f is bijective, describe its inverse. F: R?(0,?), defined by f(x) = ex Solution Integrate: f\'(x) = ?f\'\'(x)dx, from x = 0 to x = x So.... f\'(x) = -4?sin(2x)dx, from x = 0 to x = x 2cos(2x) - 2cos(2*0) = 2cos(2x) - 2 But you HAVE to add f\'(0): f\'(x) = f\'(0) + ?... = 1 + 2cos(2x) - 2 --> f\'(x) = 2cos(2x) - 1 (you could have ALSO just assumed: f\'(x) = 2cos(2x) + b, then find b based off f\'(0) = 1) So now integrate to find f(x): f(x) = f(0) + ?f\'(x)dx, from x = 0 to x = x ?f\'(x)dx = ?{2cos(2x) - 1)dx --> sin(2x) - x So, we have: sin(2x) - x - (sin(2*0) - 0) = sin(2x) - x - 0 = sin(2x) - x So we have: f(x) = f(0) + ?... = 0 + sin(2x) - x So: f(x) = sin(2x) - x Therefore: f(p/6) = sin(p/3) - p/6 = sin(60) - p/6 = v(3)/2 - p/6 .