1. The problem statement, all variables and given/known data I can do the question, but in a different way to the worked solution which I don't understand. So my question is can anyone explain the worked solution which is in point 3 below. The question was to show there is exactly one zero to the function f(x) = Ax^3 - Ax + 1, with A>0, when A < 3sqrt(3) / 2. 2. Relevant equations The part above this asks to find the stationary points, and there are two of them, at + and - 1/sqrt(3). 3. The attempt at a solution My reasoning was that since A>0, which is the coefficient of x^3 (i.e. the highest degree term), the graph goes up into the top right of the first quadrant as x and y approach positive infinity. And therefore to have only one root, the y value of the stationary point at x = 1/sqrt(3) must be >0. (I also considered the case where the stationary point at 1/sqrt(3) was <0 and the one at -1/sqrt(3) was also <0, but this is impossible given the constant term of +1 and from playing with graphing software, and having A>0). Solving this, i.e. y = f(x) being >0 at x=1/sqrt(3) gives the correct answer of A < 3sqrt(3) / 2 from point 1 above. HOWEVER, the worked solution calculates the y values at both + and - 1/sqrt(3), and then multiplies the y values together, giving a quadratic in A (i.e. containing A^2), and then uses the condition that the two y values multiplied together must be >0 for there to be only one zero. Which I don't understand at all... i.e. f(1/sqrt(3)) * f(-1/sqrt(3)) = (1 - 2A/3sqrt(3))(1 + 2A/3sqrt(3)) = 1 - 4A^2 / 27 and when this is >0, i.e. when A < 3sqrt(3)/2, there will be only one zero of f(x).