# What is the equation of the tangent line to the curve f(x)=3x^2+1, at the point x=1?

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We'll recall the fact that the derivative of a function at a given point is represented by the value of the slope of the tangent line at the curve.

The equation of the tangent line, in the point x = 1 is:

y - f(1) = f'(1)(x-1)

We'll calculate f(1), by substituting x by 1 in the expression of the function:

f(1) = 3*1^2 + 1

f(1) = 3 + 1

f(1) = 4

To calculate f'(1), first we'll have to differentiate the given function with respect to x:

f'(x) = (3x^2 + 1)'

f'(x) = 6x

Now, we'll replace x by 1 in the expression of the first derivative:

f'(1) = 6

Now, we'll substitute f(1) and f'(1) in the expression of the equation of the tangent line:

y - f(1) = f'(1)(x-1)

y - 4 = 6(x - 1)

We'll remove the brackets:

y - 4 = 6x - 6

We'll add 6 both sides:

y = 6x - 6 + 4

y = 6x - 2

**The equation of the tangent line, to the curve f(x) = 3x^2 + 1, is: y = 6x - 2.**