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!
I have a question from my recent test:
A quartic function has the equation y = (x - 2) (x - 4) (3x^2 + kx + 8), where k is any real number. How many stationary points can the quartic have?
I have tried differentiating and equating to zero. However, the cubic equation that this yields does not readily allow the number of solutions (and hence stationary points) to be deduced. The answer is apparently 1 or 3, but I do not know how to show that 2 can't be the answer too. I recognise that we don't need to necessarily find the stationary points....we simply need to find how many there are. However, I don't know how else to proceed. I would appreciate any help on how to approach this. A graphics calculator (eg. TI-89 or TI-nspire) is allowed.
I am currently advanced high school level.