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

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To find the equation of the tangent to y = 3x-x^3 at x= 2.

The tangent to any curve is given by:

y - y1 = {(dy/dx) at x= x1. }{x-x1)....(1)

Put x = x1 = 2 in 3x -x^3 , and we get y1 = 3*2-2^3 = 6-8 = -2.

So (x1 , y1) = (2, -2).

dy/dx = {3x-x^3}' = 3 -3x^2.

{(dy/dx )at x= x1= 2} = 3-3*2 = 3-6 = -3.

Substituting (x1,y1) = (2,-2) and {(dy/dx)at x=x1) = -3 in eq (1), we get:

y - (-2) = (-3) {x- 2}.

y+2 = -3x+6.

3x+y+2-6 = 0

3x+y-4 = 0 is the required tanget to y = 3x-x^3 at x= 2.

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**