Determine the equation of the tangent to the curve y=3x-x^3 at x=2 .

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You need to use the equation of tangent line to the curve, at a given point `(x_0,y_0)` , such that:

`y - y_0 = f'(x_0)(x - x_0)`

The problem provides `x_0 = 2` , hence, you may evaluate `y_0` replacing 2 for x in equation of the function, such that:

`y_0 = f(x_0) = f(2) => y_0 = 6 - 2^3 => y_0 = 6 - 8 => y_0 = -2`

You need to differentiate the function with respect to x, such that:

`f'(x) = 3 - 3x^2 => f'(x_0) = f'(2) => f'(2) = 3 - 12 = -9`

You need to replace -2 for `y_0` , -9 for `f'(x_0)` and 2 for `x_0` in tangent equation, such that:

`y + 2 = -9(x - 2) => y = -9x + 18 - 2 => y = -9x + 16`

**Hence, evaluating the equation of the tangent line to the given curve, at` x = 2` , yields **`y = -9x + 16.`

To determine the equation of the tangent line to the graph of y, we'll have to determine the derivative of y at x = 2.

f'(2) = lim [f(x) - f(2)]/(x-2)

f(2) = 3*2 - 2^3

f(2) = 6 - 8

f(2) = -2

lim [f(x) - f(2)]/(x-2) = lim (3x - x^3 + 2)/(x-2)

We'll substitute x by 2:

lim (3x - x^3 + 2)/(x-2) = (6-8+2)/(2-2) = 0/0

Since we've obtained in indeterminacy acse, we'll apply L'Hospital rule:

lim (3x - x^3 + 2)/(x-2) = lim (3x - x^3 + 2)'/(x-2)'

lim (3x - x^3 + 2)'/(x-2)' = lim (3-3x^2)/1

We'll substitute x by 2:

lim (3-3x^2)/1 = 3 - 3*4 = 3-12 = -9

f'(2) = -9

But the derivative of y at x = 2 is the slope of the tangent line to the curve y.

m = -9

Now, we'll write the equation of the tangent line, whose slope is m=-9 and it passes through the point that has x coordinate = 2.We'll compute the y coordinate of this point:

f(2) = 3x - x^3

f(2) = 6 - 8

f(2) = -2

The equation of the tangent line is:

y - (-2) = m(x - 2)

y + 2 = -9(x - 2)

We'll remove the brackets:

y + 2 = -9x + 18

y + 9x + 2 - 18 = 0

y + 9x - 16 = 0

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