A curve has an equation y=3x-2x^2-x^3. Find the x-coordinate(s) of the stationary point(s) of the curve.

The very first step in solving this problem is understanding that a stationary point is where the derivative of the curve, dy/dx (or in Newton’s notation y’), is equal to zero. This is because at stationary points the tangent to the curve is horizontal and so its gradient (the geometric equivalent to a derivative) is equal to zero.Now, we take the derivative of y w.r.t x: y’=3-4x-3x^2.Equating this to zero gives us a quadratic equation in x: 3-4x-3x^2=0 <-> 3x^2+4x-3=0.Solving this equation by completing the square, which has the formula ax^2+bx+c=0 <-> a[x+(b/2a)]^2-(b/2a)^2+c=0,will give us 3(x+2/3)^2-4/9-1=0 -> (x+2/3)^2=13/9Taking the square-root of both sides: x+2/3=±(√13)/3 -> x=(1/3)[-2±√13]Therefore, the x-coordinates of each stationary point are x=(1/3)[-2+√13] and x=(1/3)[-2-√13].