Find the x and y coordinates of the turning points of the curve 'y = x^3 - 3x^2 +4'. Identify each turning point as either a maximum or a minimum.

The first part of the problem is solved by differentiating once and equating this to zero:
y = x^3 - 3x^2 +4 .dy/dx = 3x^2 - 6x .dy/dx = x(3x - 6) .
At the turning points;
x(3x - 6) = 0 (turning points occur where the gradient, dy/dx, equals zero) .
Hence, x = 0 or 2.
Inputting these x-values into the original equation yields the respective y-coridnates or the turning points. The locations are (0, 4) and (2, 0).
The nature of the turning points can be determined by finding the second derivative of the original equation:
d^2y/dx^2 = 6x - 6 .
At (0, 4), d^2y/dx^2 = -6 .At (2, 0), d^2y/dx^2 = 6 .
Therefore (0, 4) is a maximum and (2, 0) is a minimum (positive second derivative --> minimum, negative second derivative --> maximum).

RM
Answered by Robbie M. Maths tutor

12083 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Compute the indefinite integral of x^8 ln(3x)dx


When you are working out dy/dx = 0, why do you do this and what does it mean?


How do you differentiate 2 to the power x?


Find the equation of the tangent to the curve y = (5x+4)/(3x -8) at the point (2, -7)


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