A curve has equation y = 20x - x^2 - 2x^3 . The curve has a stationary point at the point M where x = −2. Find the x- coordinate of the other stationary point of the curve
A stationary point on a curve is when the differential of the equation of the curve = 0. In other words when dy/dx = 0Take the equation in the question: y = 20x - x2 - 2x3A simple rule to find the differential of the curve is to multiply that power by the value and drop that power by one. e.g. the differential of 2x3: dy/dx = 6x2.Therefore, dy/dx = 20 - 2x - 6x2.As the stationary point is when dy/dx = 0. Therefore, 20 - 2x - 6x2 = 0.factorise this quadratic. (x + 2 )(-6x + 10) = 0Therefore, the other stationary point is x = 10/6 which can be simplified to x = 5/3
HJ
Related Maths A Level answers
All answers ▸how to sketch a funtion f(x)
How do I find the equation of the normal to the curve y=x^2 at the point (x1,y1)? Where x1=2 and y1=4 .
Find dy/dx in terms of t for the curve defined by the parametric equations: x = (t-1)^3, y = 3t - 8/t^2, where t≠0
What are partial fractions for and how do I find them?