Question

Equation of tangent to the curve \( y = x + \frac{4}{x^2} \) which is parallel to x-axis is

• A. y = 8
• B. y = 0
• C. y = 3
• D. y = 2

Hint:

A tangent parallel to the x-axis has a slope of 0.
Find \(\frac{dy}{dx}\) for the curve and set it to 0 to find the x-coordinate(s) of the point(s) of tangency.
Substitute these x-values back into the curve's equation to find the y-coordinate(s).
The equation of a horizontal line is \(y = \text{constant}\).

Answer 1

The Correct Option is C

The curve is \( y = x + \frac{4}{x^2} = x + 4x^{-2} \). We need the equation of the tangent that is parallel to the x-axis. A line parallel to the x-axis has a slope of 0. The slope of the tangent to the curve is given by its derivative \(\frac{dy}{dx}\). So, we need to find points on the curve where \(\frac{dy}{dx} = 0\). Calculate \(\frac{dy}{dx}\): \( \frac{dy}{dx} = \frac{d}{dx}(x + 4x^{-2}) = 1 + 4(-2)x^{-2-1} = 1 - 8x^{-3} = 1 - \frac{8}{x^3} \). Set the slope to 0: \( 1 - \frac{8}{x^3} = 0 \) \( 1 = \frac{8}{x^3} \) \( x^3 = 8 \) \( x = \sqrt[3]{8} = 2 \). So, the tangent is parallel to the x-axis at the point where \(x=2\). Now find the y-coordinate of this point on the curve: \( y = x + \frac{4}{x^2} \). Substitute \(x=2\): \( y = 2 + \frac{4}{2^2} = 2 + \frac{4}{4} = 2 + 1 = 3 \). The point of tangency is \((2,3)\). The tangent line is parallel to the x-axis and passes through \((2,3)\). A line parallel to the x-axis has the equation \(y = \text{constant}\). Since it passes through \((2,3)\), the equation of the tangent is \(y = 3\). This matches option (c). \[ \boxed{y=3} \]