Find the stationary points of the function y = (1/3)x^3 + (1/2)x^2 - 6x + 15

A stationary point is a point on the function where the gradient is zero. The phrase 'stationary point' coming up in a question always indicates that differentiation may be useful to solve it. In this case, the derivative of the function, often expressed as dy/dx, is x^2 + x - 6. As dy/dx is the gradient of the function, set it equal to zero to find stationary points. The easiest way to solve x^2 + x - 6 = 0 is by factorisation. So (x+3)(x-2)=0 gives the solutions x=2 , x=-3. Sub these back in to the original equation to find the corresponding y values. For x=2, y=23/3. For x=-3, y=57/2. The stationary points are therefore at (2, 23/3) and (-3,57/2).

Answered by Matthew H. Maths tutor

8883 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given (x-2) is a factor of ax^3 + ax^2 + ax - 42, find the value of a


Let f(x) = 2x^3 + x^2 - 5x + c. Given that f(1) = 0 find the values of c.


Find the tangent to the curve y = x^3 - 2x at the point (2, 4). Give your answer in the form ax + by + c = 0, where a, b and c are integers.


What's the gradient of the curve y=x^3+2x^2 at the point where x=2?


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