y=f(x)=3x³-3x ··· Eq.1
Since the given curve is continuous function, Eq.1 is differetiable at every x and the first derivative f'(x), the slope (tangent) of the original curve, is found at every point (x,y) on the curve.
The 1st derivative of Eq.1 is: f'(x)=9x²-3, so the slope of the tangent at x=1 is: f'(1)=9(1)²-3=6
While, the equation of a line that passes thru a point (x1,y1), and has a slope of m is:
(y-y1)=m(x-x1) ··· Eq.2 (the point- slope form of linear function)
Here, m=f'(x1)=6, x1=1,y1=0 Plug these values into Eq.2. Eq.2 can be restated and simplified as follows: y-0=6(x-1) ⇒ y=6x-6
Therefore, the tangent to the given curve at point (1,0) is: y=6x-6