Find the turning points of the curve y=2x^3 - 3x^2 - 14.

First differentiate the equation: dy/dx = 6x^2 - 6xSet this equal to 0 as at turning points the change in gradient is 0: 0 = 6x^2 - 6x6x(x-1)=06x=0 therefore x=0(x-1)=0 therefore x=1x=1,0Now substitute back into equation 1 to find y values and hence co ordinatesx=1 y=-15x=0 y=-14

ER
Answered by Edward R. Maths tutor

5350 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find f'(x) and f''(x) when f(x) = 3x^2 +7x - 3


Differentiate y = 4ln(x)x^2


Differentiate y = 5x^3 + 7x + 3 with respect to x


Integrate sinx*ln(cosx) with respect to x.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning