At what points does the curve defined by y = x^3 – 3x + 3 have horizontal tangents.

To find the points at which the curve y = x^3 – 3x + 3
have the horizontal tangents.

If the tangents are
horizontal, then the slope dy/dx at that point is
zero.

=> dy/dx = (x^3-3x+3)' =
0.

=> 3x^2-3 =
0.

=> 3x^2=
3.

=> x^2 = 3/3 =
1.

=> x1= sqrt1, or x2=
-sqrt1.

So x1=1 , or x2 =
-1.

Put x= x1  in y= x^3-3x+3 to get
y1.

When x= x1 = 1, y1 = 1^3-3*1+3 =
1.

When x= x2 = -1, y2 = (-1)^3-3(-1)+3 =
5.

Therefore (x1,y1) = (1,1)
and (x2,y2) = (-1, 5) , dy/dx = 0 and the tangents are ||
to x axis or horizontal.