Let f(x) = x^3 + 3x^2 + 6x + 4. Find the interval in which f^-1(x) exists, has derivate and is positive
Correct: C
f'(x) = 3x^2+6x+6 = 3(x^2+2x+2) = 3((x+1)^2 + 1)>0, so f(x) is strictly increasing. f''(x) = 6x+6, so f''(x)>0 when x>-1. f(-1) = 1-3+6 =4+1=2. So f^-1(x) exists and is positive between [2, infinity). Therefore, f^-1(x) exists, derivative at range (-1, infinity)