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Find the tangent to the curve y=x^3+3 at the point x=1.

1. Differentiate the Equation of the curve to find the gradient: y'=3x^2

2. The gradient of the tangent is found by substituting x=1 into y'=3x^2: Gradient of tangent=3

3. Now we must find out the co-ordinates of the point. These are (x=1y=1^3+3) = (1,4).

4. Now to find out the equation of the tangent, substitute x=1, y=4 and m=3 into y=mx+c to get c=1, (the y-intercept)

5. This gives the tangential equation as y=3x+1.

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