To find the equation of the other tangent parallel to $y = x - 2$, we need to follow these steps:

Step 1: Find the slope of the given tangent line

The given tangent line is $y = x - 2$, which is a straight line. From this, we can see that the slope of the line is $1$ (since the coefficient of $x$ is $1$).

Step 2: Find the derivative of the curve to find the slope of the tangent

The equation of the curve is given by $y = 4x - x^3$. To find the slope of the tangent to the curve at any point, we need to take the derivative of the equation with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(4x - x^3) = 4 - 3x^2$

This is the slope of the tangent to the curve at any point $x$.

Step 3: Set the derivative equal to the slope of the tangent line

We are looking for the points where the slope of the tangent to the curve is equal to the slope of the given tangent line, which is $1$. Therefore, we set the derivative equal to $1$:

$4 - 3x^2 = 1$

Step 4: Solve for $x$

Now, we solve for $x$:

$4 - 3x^2 = 1$ $3x^2 = 3$ $x^2 = 1$ $x = \pm 1$

Step 5: Find the corresponding $y$-coordinates

We now substitute $x = 1$ and $x = -1$ into the original equation of the curve $y = 4x - x^3$ to find the corresponding $y$-coordinates.

1. For $x = 1$: $y = 4(1) - (1)^3 = 4 - 1 = 3$

2. For $x = -1$: $y = 4(-1) - (-1)^3 = -4 + 1 = -3$

So, the two points on the curve where the slope of the tangent is $1$ are $(1, 3)$ and $(-1, -3)$.

Step 6: Find the equation of the tangents at these points

Now that we know the slope of the tangent at these points is $1$, we can use the point-slope form of the equation of a line to find the equations of the tangents.

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is a point on the line.

1. For the point $(1, 3)$: $y - 3 = 1(x - 1)$ $y - 3 = x - 1$ $y = x + 2$

2. For the point $(-1, -3)$: $y + 3 = 1(x + 1)$ $y + 3 = x + 1$ $y = x - 2$

Step 7: Conclusion

The two tangents to the curve $y = 4x - x^3$ that are parallel to the line $y = x - 2$ are:

1. $y = x + 2$
2. $y = x - 2$ (this is the given tangent line)

Thus, the equation of the other tangent parallel to $y = x - 2$ is $y = x + 2$.

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Would you like more details on any of the steps? Here are some follow-up questions:

1. How do we interpret the slope of a curve geometrically?
2. Can we find tangents to a curve with higher-order polynomials?
3. How do we determine if a point is a maximum or minimum using the first derivative?
4. What is the significance of the second derivative in curve analysis?
5. How do we find tangents to implicit functions?

Tip: When solving for tangents, always start by computing the derivative of the curve, as it gives the slope of the tangent at any point on the curve.