A curve has the equation y=3 + x^2 -2x^3. Find the two stationary points of this curve.

At stationary point the derivative of y with respect to x equals zero. Find dy/dx. dy/dx = 2x - 6x2.Set dy/dx = 0 ,giving 2x - 6x2 = 0. We can rearrange this to x(1-3x) = 0 and solve the equation for x. This results in x1 = 0 and x2 = 1/3.Substitute these x values into the original equation and we get the corresponding y values for the coordinates. y1 = 3 + (0)2 - 2(0)3 = 3. y2 = 3 +(1/3)2 - 2(1/3)3 = 82/27. Giving the coordinates for the two stationary points as p1= (0,3) and p2 = (1/3,82/27).

TC
Answered by Thomas C. Maths tutor

4228 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Derive the quadratic equation.


a) show that (cosx)^2=8(sinx)^2-6sinx can be written as (3sinx-1)^2=2 b)Solve (cosx)^2=8(sinx)^2-6sinx


What is the area bound by the x-axis, the lines x=1 and x=3 and the curve y=3x^(2)-1/x ? Answer in exact form.


What the integral of e^2x*x? (limits 0,1)


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