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 ▸

Express (7+ √5)/(3+√5) in the form a + b √5, where a and b are integers.


Find any stationary points in the function f(x) = 3x^2 + 2x


Find the stationary points of y=x^3 + 3x^2 - 9x - 4


The equation of a curve is y = x^2 - 5x. Work out dy/dx


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