Find the coordinates of the turning points of the curve y = 4/3 x^3 + 3x^2-4x+1

First differentiating by the rule that xn differentiates to nxn-1 we have that dy/dx = 4x2+6x-4.
At the turning points of a curve the differential is equal to 0 so we set 0=dy/dx = 4x2+6x-4, by factorising we can see that 0= 2(2x2+3x-2) = 2(2x-1)(x+2), so our turning points are at x=1/2 and x=-2 as this is when dy/dx =0.
to find the coordinates for these points we plug the x values into the original equation y=4/3 x3+3x2-4x+1 and we find that the turning points are (1/2, -1/12) and (-2, 31/3).

TE
Answered by Theo E. Maths tutor

8418 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate (lnx)/x^2 dx between limits 1 and 5


How to expand squared brackets?


Prove that 1/(tanx) + tanx = 1/sinxcosx


Why do we have to add the +c when integrating a function


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