<p>Find the range of values of <em>a</em> for which the function
\(f(x) = x^3 + (a+2)x^2 + 3ax + 5\) is monotonic in \(\mathbb{R}\). Hence, find the set of values of <em>a</em> for which \(f(x)\) is invertible.</p>
Step-by-Step Solution
Key Concept: A function is monotonic on ℝ if and only if f'(x) has a constant sign (always ≥0 or always ≤0). For a cubic with positive leading coefficient, this requires f'(x) ≥ 0 for all x, which means the discriminant of f'(x) must be ≤0.
<p><strong>Step 1:</strong> Find f'(x).</p><p>f(x) = x³ + (a+2)x² + 3ax + 5</p><p>f'(x) = 3x² + 2(a+2)x + 3a</p><p><strong>Step 2:</strong> For monotonicity on ℝ, f'(x) must maintain constant sign for all x ∈ ℝ.</p><p>Since the coefficient of x² is positive (3 > 0), we need f'(x) ≥ 0 for all x ∈ ℝ.</p><p><strong>Step 3:</strong> Apply discriminant condition.</p><p>For f'(x) ≥ 0 for all x, the discriminant of the quadratic f'(x) must be ≤ 0:</p><p>Δ = [2(a+2)]² - 4(3)(3a) ≤ 0</p><p>4(a+2)² - 36a ≤ 0</p><p>(a+2)² - 9a ≤ 0</p><p>a² + 4a + 4 - 9a ≤ 0</p><p>a² - 5a + 4 ≤ 0</p><p><strong>Step 4:</strong> Solve the inequality.</p><p>a² - 5a + 4 = (a - 1)(a - 4)</p><p>(a - 1)(a - 4) ≤ 0</p><p>This holds when 1 ≤ a ≤ 4</p><p><strong>Step 5:</strong> Monotonic functions are bijective (one-to-one and onto), hence invertible.</p><p>∴ f(x) is monotonic and invertible when <strong>1 ≤ a ≤ 4</strong></p><p><em>Note: If the question requires strictly monotonic (strict inequality in original), then 1 < a < 4.</em></p>
Correct Answer: 1 < a < 4