If a curve has equation y = (-8/3)x^3 - 2x^2 + 4x + 18, find the two x coordinates of the stationary points of this curve.
The first step in solving this question is to differentiate equation y with respect to x. To differentiate a given 'x' term in an equation use the following method --> xn becomes nx(n-1).y = (-8/3)x3 - 2x2 + 4x + 18dy/dx = -8x2 - 4x + 4A stationary point is found when dy/dx is equal to zero. This means dy/dx can now be solved for x using the quadratic formula. x = (- b +/- √(b2 - 4ac))/2aUsing this formula and identifying a = -8, b = -4 and c = 4, x = (4 +/- √(16 + 128))/ - 16 = (4 +/- 12)/ - 16the two x coordinates of the stationary points of the curve y can be found to be:x = (4 + 12)/ - 16 = -1and x = (4 - 12)/ - 16 = 1/2
RT
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