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.
dy/dx = 20 - 2x - 6x2 = -6x2 - 2x + 20 which factorises to dy/dx = (10 - 6x)(2 + x). This shows that x=-2 and x=10/6=5/3 are stationary points of y = 20x −x2 −2x3.
XY
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A curve has equation y = 20x −x2 −2x3 . (A) Find the x-coordinates of the stationary points of the curve.