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(B) Given the curve equation: $y = x + \frac{4}{x^{2}}$.To find the slope of the tangent,we differentiate with respect to $x$:$\frac{dy}{dx} = 1 - \fr

Vedclass · Accepted Answer

$y=3$

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(B) Given the curve equation: $y = x + \frac{4}{x^{2}}$.To find the slope of the tangent,we differentiate 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 $0$:$1 - \frac{8}{x^{3}} = 0$.Solving for $x$:$\frac{8}{x^{3}} = 1 \Rightarrow x^{3} = 8 \Rightarrow x = 2$.Now,find the corresponding $y$-coordinate by substituting $x = 2$ into the original curve equation:$y = 2 + \frac{4}{2^{2}} = 2 + \frac{4}{4} = 2 + 1 = 3$.The point of contact is $(2, 3)$.Since the tangent is parallel to the $x$-axis,its equation is of the form $y = k$. Substituting the $y$-coordinate of the point of contact,we get $y = 3$.

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

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