If a curve has equation y = (-8/3)x^3 - 2x^2 + 4x + 18, find the two x coordinates of the stationary points of this curve.

The first step in solving this question is to differentiate equation y with respect to x. To differentiate a given 'x' term in an equation use the following method --> xn becomes nx(n-1).y = (-8/3)x3 - 2x2 + 4x + 18dy/dx = -8x2 - 4x + 4A stationary point is found when dy/dx is equal to zero. This means dy/dx can now be solved for x using the quadratic formula. x = (- b +/- √(b2 - 4ac))/2aUsing this formula and identifying a = -8, b = -4 and c = 4, x = (4 +/- √(16 + 128))/ - 16 = (4 +/- 12)/ - 16the two x coordinates of the stationary points of the curve y can be found to be:x = (4 + 12)/ - 16 = -1and x = (4 - 12)/ - 16 = 1/2

RT
Answered by Robyn T. Maths tutor

2888 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

I don't understand why the function "f(x)=x^2 for all real values of x" has no inverse. Isn't sqrt(x) the inverse?


How do I integrate ln(x), using integration by parts?


The velocity of a moving body is given by an equation v = 30 - 6t, where v - velocity in m/s, t - time in s. A) What is the acceleration a in m/s^2? B) Find the expression for the displacement s in terms of t given the initial displacement s(0)=10 m.


Determine the integral: ∫x^(3/4)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