For the curve f(x) = 2x^3 - 54x, find the stationary points and state the nature of these points

Firstly, find the values of x where f'(x) = 0

f'(x) = 6x2 - 54

6x2 - 54 = 0

6(x+3)(x-3) = 0

x = 3, y = -108 and x = -3, y = 108

Next, find the values of f''(x) at these points

f''(x) = 12x

When x = 3, f''(x) = 36 which is positive and therefore (3,-108) is a minima.

When x = -3, f''(x) = -36 which is negetive and therefroe (-3,108) is a maxima.

Answered by Ruby W. Maths tutor

4067 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

I don't understand how to visualise differentiation, please could you show my an example to allow me to understand what it actually is better?


a) Express 4(cosec^2(2x)) - (cosec^2(x)) in terms of sin(x) and cos (x) and hence b) show that 4(cosec^2(2x)) - (cosec^2(x)) = sec^2(x)


How can I understand eigenvalues and eigenvectors?


The curve C has the equation: y=3x^2*(x+2)^6 Find dy/dx


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