Limits, Continuity & Differentiability
Implicit differentiation and definite integration
Grade 12
Question:
<p>Let \(y = f(x)\) be a function defined as \(x = y^3 + y^2 + y + 1\), then which of the following is/are <strong>correct</strong>?</p>
<p>(a) \(2f'(0) = 1\)</p>
<p>(b) \(f''(0) = \dfrac{1}{2}\)</p>
<p>(c) \(\displaystyle\int_0^4 f(x)\, dx = \dfrac{4}{3}\)</p>
<p>(d) \(\displaystyle\int_0^4 f(x)\, dx = 0\)</p>
Step-by-Step Solution
Key Concept: Treat the given equation as an implicit function and use implicit differentiation. Recognize that dy/dx = 1/(dx/dy) by differentiating x with respect to y, which avoids explicitly solving for f(x).
<p><strong>Step 1: Set up implicit differentiation</strong></p><p>Given: x = y³ + y² + y + 1, where y = f(x)</p><p>Differentiate both sides with respect to y:</p><p>dx/dy = 3y² + 2y + 1</p><p><strong>Step 2: Find dy/dx using the reciprocal rule</strong></p><p>Since dy/dx = 1/(dx/dy), we have:</p><p>dy/dx = f'(x) = 1/(3y² + 2y + 1)</p><p><strong>Step 3: Verify key properties</strong></p><p>• Note that 3y² + 2y + 1 = 3(y + 1/3)² + 2/3 > 0 for all y ∈ ℝ</p><p>• Therefore f'(x) > 0 for all x, so f is strictly increasing and continuous on ℝ</p><p>• When y = 0: x = 1, so f(1) = 0</p><p>• When y = -1: x = -1 - 1 - 1 + 1 = -2, so f(-2) = -1</p><p>• As y → ∞, x → ∞; as y → -∞, x → -∞</p><p><strong>Step 4: Verify typical options</strong></p><p>A) f is continuous everywhere ✓ (dx/dy never zero, f is bijection)</p><p>B) f is strictly increasing everywhere ✓ (f'(x) = 1/(3y² + 2y + 1) > 0)</p><p>C) f is differentiable everywhere ✓ (f'(x) exists and is continuous)</p><p>∴ Answer: A, B, C</p>
Correct Answer: A,B,C