# At what points, if any, does the graph of y= x^3 – 3x + 3 have horizontal tangents?

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y = x^3-3x+3.

The horizontal tangent should have slope zero. Or dy/dx = 0.

So wefind dy/dx and equate to zero.

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

dy/dx = 3x^2-3.

dy/dx = 0 gives:

3x^2-3 = 0

x^2-1 = 0

(x-1)(x+1) = 0

Therefore x-1 = 0. Or x = - 1 for which which dy/dx = 0.

The corresponding y values:

When x= 1, y = 1^3-3*1+3 = 1

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

Therefore at (1 ,1) and (-1,5) the curve y = x^3-3x+3 has horizontal tangents. y =1 and y =5

To find horizontal tangents we have to look for points where the slope is zero. To do that, we set the formula for the slope equal to zero and solve.

Now as y = x^3 – 3x + 3, the slope is equal to the derivative of x^3 – 3x + 3 which is equal to 3x^2 -3.

So 3x^2 – 3 =0

=> x^2 – 1 =0

=> x^2 = 1

=> x = 1 or x =-1.

Therefore the x- coordinates of the points we are looking for are 1 and -1. For the y coordinates,

when x= 1, y = 1^3 – 3*1 +3 =1

when x = -1, y = -1^3 – 3*-1 +3 = 5

**The two points where the tangents are horizontal are (1, 1) and (-1, 5)**