Math Is Fun Forum

thank you for your reply, bob. i am still confused because of the effect of the k, which prevents us from solving the cubic to get three solutions. so far i have:

dy/dx = 12x^3 + (3k-54)x^2 + (64-12k)x + (8K - 48)

second derivative = 36x^2 + 6(k-18)x +64 - 12k

For this second derivative, b^2 - 4ac = 36k^2+432k+2448, which is always positive since discriminant <0 for this quadratic involving k and the quadratic involving k is an upright parabola.

So, this means that we have two values at which the second derivative equals zero. However, i do not know how to check if these coincide with stationary points.

in short, with the original quartic, there must be a stationary point between x=2 and x=4. How do we work out whether there could be a stationary point of inflection for x<2?? my issue is still proving whether or not we can have 2 stationary points for the quartic graph

thanks again for your help everyone!