Find the x-coordinates of the turning points of the function f(x)=x^3+2x^2-9x-18, where y=f(x).
The turning points of a function are always where the gradient at that point is 0, so setting the derivative of the function equal to 0 will give the turning points.dy/dx or df/dx of the function will yield df/dx=3x2+4x-9.Set df/dx=0 as gradient is 0.Then solve the quadratic 3x2+4x-9=0(solve using quadratic formula) where a=3, b=4, c=-9x=-2/3+Sqrt(31)/3 or -2/3-Sqrt(31)/3
