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

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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).**

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).**

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.