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WilsonWorla
find all points on the curve y=x^3-3X where the tangent is parallel to the x-axis.
Take the derivative \[y=3x^2-3\] \[y(0)=3x^2-3\] when \[x=1,x=-1\] Now find y's for x=1,-1. Tangent is parallel to the x-axis at these points.
When in doubt always start with the derivative of the function. Hence if y = x³ − 3x dy/dx = 3x² − 3 where dy/dx is of course the instantaneous rate of change function (the slope function). Since the tangent to y is parallel to the x-axis, dy/dx = 0 ∵ the slope of the x-axis is zero (x-axis is horizontal and the slope of any horizontal line is zero). Thus dy/dx = 0 and since dy/dx = 3x² − 3, 3x² − 3 = 0 solving for x, we obtain x = ±1 when x = -1, y = 2 when x = 1, y = -2 ∴ the tangent to y = 3x² − 3 is parallel to the x-axis at the points (-1, 2) and (1, -2)