I'm a beginning calculus student and this is one of the 'challenge' questions from an older edition of the Stewart's Calculus series of textbooks. I believe I did the question properly, but the solution in the book is completely different from my own attempt. I'm not good enough at math yet to feel confident that my solution has no logical flaws or unwarranted assumptions. I would very much appreciate it if someone could have a look at my solution and let me know if it's mathematically sound. 1. The problem statement, all variables and given/known data Let P be a point on the curve y = x^3 and suppose the tangent line at P intersects the curve again at Q. Prove that the slope of the tangent at Q is four times the slope of the tangent at P. 3. The attempt at a solution Let the x-coordinate of P be "a", and let the x-coordinate of Q be "b". Also, let the slope of the tangent at P be "mL", and let the slope of the tangent at Q be "mK". 1. First I considered the tangent at P to be a secant line since it intersects the graph at two points. Using the formula for the slope of the secant, (f(x + h) - f(x))/h, I got: (f(b + h) - f(b)) / h = mL 3b^2 + 3bh + h^2 = mL And, since h is the distance from a to b, h = |a - b|. Consider that a > b, and so |a - b| = a - b. For the case in which b > a, the slope of the secant becomes 3a^2 + 3ah + h^2, and h = b - a. The leaves the final solution unaffected. Also, mL = 3a^2, since that is the value of the derivative at a. Substituting these values into the above equation and simplifying yields: b^2 + ab - 2a^2 = 0 Solving for b using the quadratic formula yields: b = a, which can be discarded b = -2a Since b = -2a: f'(b) = 3b^2 f'(b) = 3(-2a)^2 f'(b) = 12a^2 And since f'(a) = 3a^2, the slope of the tangent at Q is four times the slope of the tangent at P. Correct?