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Question

(C) Given the curve equation: $y = x + \frac{4}{x^2}$.Find the derivative with respect to $x$: $\frac{dy}{dx} = 1 - \frac{8}{x^3}$.Since the tangent i

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$y = 3$

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(C) Given the curve equation: $y = x + \frac{4}{x^2}$.Find the derivative with respect to $x$: $\frac{dy}{dx} = 1 - \frac{8}{x^3}$.Since the tangent is parallel to the $X$-axis,its slope must be $0$. Therefore,set $\frac{dy}{dx} = 0$.$1 - \frac{8}{x^3} = 0 \Rightarrow \frac{8}{x^3} = 1 \Rightarrow x^3 = 8 \Rightarrow x = 2$.Now,find the corresponding $y$-coordinate by substituting $x = 2$ into the original curve equation:$y = 2 + \frac{4}{2^2} = 2 + \frac{4}{4} = 2 + 1 = 3$.Since the tangent is a horizontal line passing through $(2, 3)$,its equation is $y = 3$.

The equation of the tangent to the curve $y = x + \frac{4}{x^2}$ which is parallel to the $X$-axis is:

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