<p>Let \(f(x) = 30 - 2x - x^3\). If \(f(f(f(x))) > f(f(-x))\), find the highest integral value of \(x\) satisfying this inequality.</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = 30 - 2x - x³ is a strictly decreasing function (f'(x) = -2 - 3x² < 0 for all x). Therefore, composing f preserves monotonicity, and f(f(f(x))) > f(f(-x)) simplifies to an inequality by applying the decreasing property repeatedly.
<p><strong>Step 1:</strong> Find f'(x).</p><p>f'(x) = -2 - 3x² < 0 for all x ∈ ℝ</p><p>Therefore, f is strictly decreasing on ℝ.</p><p><strong>Step 2:</strong> Apply the decreasing property to f(f(f(x))) > f(f(-x)).</p><p>Since f is strictly decreasing: f(A) > f(B) ⟺ A < B</p><p>Apply once: f(f(x)) < f(-x)</p><p>Apply again: f(x) > -x</p><p><strong>Step 3:</strong> Solve f(x) > -x.</p><p>30 - 2x - x³ > -x</p><p>30 - x - x³ > 0</p><p>x³ + x - 30 < 0</p><p><strong>Step 4:</strong> Find where x³ + x - 30 = 0.</p><p>Testing x = 3: 27 + 3 - 30 = 0 ✓</p><p>Factor: (x - 3)(x² + 3x + 10) = 0</p><p>The quadratic x² + 3x + 10 has discriminant 9 - 40 = -31 < 0, so only x = 3 is a real root.</p><p><strong>Step 5:</strong> Determine the sign of x³ + x - 30.</p><p>Since the coefficient of x³ is positive, x³ + x - 30 < 0 when x < 3.</p><p>The highest integral value satisfying x < 3 is x = 2.</p><p>∴ Answer: 2</p>
Correct Answer: 2