Answer:

The derivative of f(x) is f'(x) = 9x^2 - 2. To solve for the tangent line using the point-slope formula, you need to find the point where the tangent line to f(x) at the point (x, f(x)) intersects the line defined by the equation D(x).

The intersection point of the line and f(x) at the point (x, f(x)) will be (x, f(x) + D(x)) = (x, f(x) + 3x + 2).

So, the slope of the tangent line to f(x) at the point (x, f(x)) will be:

m = (f(x) + 3x + 2) - f(x) / x - x

m = -2 / -1

m = 2

For the tangent line to pass through the point (x, f(x) + D(x)), we would need:

y - f(x) - D(x) = 2x + 2

Expanding the second equation:

y = 3x^3 - 2x^2 + x + 3 - x - 2

Then, we need to set the above equation equal to the tangent line:

y = 2x + k

So, we have:

2x + k = 3x^3 - 2x^2 + x + 3 - x - 2

Expand both sides and simplify:

2x + k = 3x^3 - 2x^2 + x + 1

Collecting like terms:

2x + k = x^3 + x

Setting everything on one side and removing the x terms:

k = x^3 - 1

So, the tangent line to the curve f(x) is given by the equation:

y = 2x + f(x)^3 - 1

Therefore, the tangent line to the curve f(x) at the point (x, f(x)) is given by the equation:

y = 2x + f(x)^3 - 1.