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 

Answered by Jan G. MAT tutor

4509 Views

See similar MAT University tutors

Related MAT University answers

All answers ▸

How many 0's are at the end of 100! (100 factorial)?


If f(x) =x^2 - 5x + 7 what are the coordinates of the minimum of f(x-2)?


What is the square root of the imaginary number i?


Can you please help with Question 5 on the 2008 MAT?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy