Applications of Derivatives
Tangents to curves
Grade None

Question:

<p>The equation of the tangent to the curve \(y = x + \dfrac{4}{x^2}\), that is parallel to the x-axis, is</p>
<p>\(y = 1\)</p>
<p>\(y = 2\)</p>
<p>\(y = 3\)</p>
<p>\(y = 0\)</p>

Step-by-Step Solution

Key Concept: A tangent parallel to the x-axis has slope zero, so we must find where dy/dx = 0, then determine the y-coordinate at that point to get the equation.
<p><strong>Step 1:</strong> Find the derivative of y = x + 4/x²</p><p>dy/dx = 1 + 4·(-2)x⁻³ = 1 - 8/x³</p><p><strong>Step 2:</strong> For tangent parallel to x-axis, set dy/dx = 0</p><p>1 - 8/x³ = 0</p><p>8/x³ = 1</p><p>x³ = 8</p><p>x = 2</p><p><strong>Step 3:</strong> Find y-coordinate when x = 2</p><p>y = 2 + 4/2² = 2 + 1 = 3</p><p><strong>Step 4:</strong> Equation of horizontal line through (2, 3)</p><p>y = 3</p><p>∴ Answer: C</p>
Correct Answer: C

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