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

5538 Views

See similar MAT University tutors

Related MAT University answers

All answers ▸

When is the inequality x^4 < 8x^2 + 9 true?


What graph can y = cos^2(x^2)/ x^2 have, for x > 0 ?


How many distinct solutions does the following equation have? log(base x^2 +2) (4-5x^2 - 6x^3) = 2 a)None, b)1, c)2, d)4, e)Infinitely many


What is the square root of the imaginary number i?


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences