The horizontal tangent is the line where the slope is 0.

Then we will find the values of x such that the first derivative of the function is 0.

==> y = x^3 - 3x + 3

Let us differentiate.

==> y' = 3x^2 - 3 = 0

==> 3x^2...

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The horizontal tangent is the line where the slope is 0.

Then we will find the values of x such that the first derivative of the function is 0.

==> y = x^3 - 3x + 3

Let us differentiate.

==> y' = 3x^2 - 3 = 0

==> 3x^2 = 3

==> x^2 = 1

==> x1 = 1 ==> y1= 1-3+3 = 1

==> x2= -1.==> -1 +3 + 3 = 5

Then there are two horizontal tangent lines for the curve y.

** The horizontal tangent lines are at the points ( 1,1) and (-1, 5).**

Horizontal lines have a slope of 0. Therefore to find the points where the curve y = x^3 – 3x + 3 has horizontal tangents we need to find the points where the slope of the tangent is 0.

The slope of the tangent to a curve is given by the first derivative of the curve.

For y = x^3 – 3x + 3

y’ = 3x^2 - 3

Equating this to 0

=> 3x^2 – 3 = 0

=> x^2 – 1 =0

=> x^2 = 1

=> x = 1 or -1

At x = 1, y = 1 – 3 +3 = 1

At x = -1, y = -1 + 3 +3 = 5

**So the required points where the tangent to the curve are horizontal lines are (1, 1) and (-1, 5).**