Find the coordinates of the stationary points on the curve y=x^5 -15x^3
dy/dx = 5x4-45x2 by 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)
SD
Answered by Shavon D. • Further Mathematics tutor
5887 Views
See similar Further Mathematics GCSE tutorsRelated Further Mathematics GCSE answers
All answers ▸If y=x^3+9x, find gradient of the tangent at (2,1).
find the stationary point of the curve for the equation y=x^2 + 3x + 4
Why does the discriminant b^2-4ac determine the number of roots of the quadratic equation ax^2+bx+c=0?
The function f is given by f(x) = SQRT(2x − 5). Work out x when f(x) = 1.2