Key Concepts and Formulas

• The derivative of a function, $\frac{dy}{dx}$, represents the slope of the tangent line to the curve $y=f(x)$ at a given point.
• A line parallel to the x-axis is a horizontal line, and its slope is zero.
• Power Rule of Differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$.

Step-by-Step Solution

Step 1: Find the derivative of the curve

We are given the equation of the curve: $y = x + \frac{4}{x^2}$

To make differentiation easier, rewrite the second term with a negative exponent: $y = x + 4x^{-2}$

Now, differentiate $y$ with respect to $x$ using the power rule: $\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(4x^{-2})$ $\frac{dy}{dx} = 1 \cdot x^{1-1} + 4 \cdot (-2)x^{-2-1}$ $\frac{dy}{dx} = 1 - 8x^{-3}$ $\frac{dy}{dx} = 1 - \frac{8}{x^3}$

This expression represents the slope of the tangent line to the curve at any point $(x,y)$.

Step 2: Set the derivative to zero and solve for x

Since we want a tangent line parallel to the x-axis, we need to find where the slope of the tangent is zero: $\frac{dy}{dx} = 0$ $1 - \frac{8}{x^3} = 0$

Solve for $x$: $1 = \frac{8}{x^3}$ $x^3 = 8$ $x = \sqrt[3]{8}$ $x = 2$

This value $x=2$ is the x-coordinate of the point on the curve where the tangent line is parallel to the x-axis.

Step 3: Find the corresponding y-coordinate

To find the y-coordinate of the point of tangency, substitute $x=2$ back into the original equation of the curve: $y = x + \frac{4}{x^2}$ $y = 2 + \frac{4}{(2)^2}$ $y = 2 + \frac{4}{4}$ $y = 2 + 1$ $y = 3$

So, the point of tangency is $(2, 3)$.

Step 4: Write the equation of the tangent line

We have the point of tangency $(2, 3)$ and the slope $m = 0$. Since the tangent line is horizontal and passes through the point (2,3), its equation is simply $y = 3$.

Common Mistakes & Tips

• Substitute into Original Equation: Always substitute the x-value back into the original equation to find the corresponding y-value. Substituting into the derivative will give you the slope, not the y-coordinate of the point on the curve.
• Domain Considerations: Be mindful of the domain of the original function. In this case, $x \neq 0$, and our solution $x=2$ satisfies this condition.
• Recognize Horizontal Lines: Tangent lines parallel to the x-axis are horizontal lines with the equation $y = c$, where $c$ is a constant.

Summary

To find the equation of the tangent to the curve that is parallel to the x-axis, we first find the derivative of the curve. We then set the derivative equal to zero and solve for x. This gives us the x-coordinate of the point where the tangent is horizontal. We substitute this x-value back into the original equation to find the corresponding y-coordinate. Finally, we write the equation of the horizontal tangent line as $y = y_0$, where $y_0$ is the y-coordinate of the point of tangency, which yields $y=3$.

The final answer is $\boxed{3}$, which corresponds to option (C).