Let f(x) = 2x^3 − kx^2 + 2x − k. For what values of the real number k does the graph y = f(x) have two distinct real stationary points? (MAT 2017 q1.A)

A point a is a stationary point if and only if f'(a) = 0.
Let's compute the derivative:
f'(x) = (2x^3 − kx^2 + 2x − k)' = 6x^2 - 2kx + 2.  (-k is a constant so it cancells out)
f'(x) = 0 iff 6x^2 -2kx + 2 = 0
We want the equation above to have two different real roots. These exist if and only if the discriminant is strictly positive, i.e. :
(-2k)^2 - 462 > 0
4k^2 > 4*12
k^2 > 12
|k| > 2√3
Hence, our solution is: k > 2√3 or k < -2√3 

JG
Answered by Jan G. MAT tutor

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