Find the coordinates of the stationary points on the curve y=x^5 -15x^3

dy/dx = 5x4-45xby multiplying each term by the power and then decreasing the power by oneAt stationary points, dy/dx=0 since the function is neither increasing nor decreasing at a stationary point5x4-45x2=05x2(x2-9)=05x2(x-3)(x+3)=0 (Difference of two squares)Stationary points at x=0, x=3 and x=-3Plug each value into the original equation to get y coordinatesGet (0,0), (3, -162), (-3, 162)

Related Further Mathematics GCSE answers

All answers ▸

How do I find the limit as x-->infinity of (4x^2+5)/(x^2-6)?


Point A lies on the curve: y=x^2+5*x+8. The x-coordinate of A is -4. What is the equation of the normal to the curve at A?


Given y=x^3-x^2+6x-1, use diffferentiation to find the gradient of the normal at (1,5).


If the equation of a curve is x^2 + 9x + 8 = y, then differentiate it.


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