Question

find all points on the curve y=x^3-3X where the tangent is parallel to the x-axis.

Answer 1

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.

Answer 2

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)