<p>If \(a, b \in \mathbb{R}\) be fixed positive numbers such that
\(f(a + x) = b + [b^3 + 1 - 3b^2 f(x) + 3b\{f(x)\}^2 - \{f(x)\}^3]^{1/3}\)
for all \(x \in \mathbb{R}\), then prove that \(f(x)\) is a periodic function.</p>
Step-by-Step Solution
Key Concept: Recognize that the expression inside the cube root follows the binomial expansion pattern (b³ - 3b²y + 3by² - y³), which equals (b - y)³. This allows us to simplify the functional equation and establish a recursive relationship that reveals periodicity.
<p><strong>Step 1:</strong> Recognize the pattern in the expression under the cube root. We have:</p><p>b³ + 1 - 3b²f(x) + 3b{f(x)}² - {f(x)}³</p><p>Rearrange as: b³ - 3b²f(x) + 3b{f(x)}² - {f(x)}³ + 1</p><p>Notice that b³ - 3b²f(x) + 3b{f(x)}² - {f(x)}³ = [b - f(x)]³</p></p><p><strong>Step 2:</strong> Rewrite the functional equation:</p><p>f(a + x) = b + [b - f(x)]³ + 1]^(1/3)</p><p>However, more carefully: b³ - 3b²f(x) + 3b{f(x)}² - {f(x)}³ = (b - f(x))³</p><p>So the expression becomes: (b - f(x))³ + 1</p><p>Thus: f(a + x) = b + [(b - f(x))³ + 1]^(1/3)</p></p><p><strong>Step 3:</strong> Let g(x) = f(x) - b. Then f(x) = g(x) + b:</p><p>g(a + x) + b = b + [(-g(x))³ + 1]^(1/3)</p><p>g(a + x) = [1 - g(x)³]^(1/3)</p></p><p><strong>Step 4:</strong> Apply the functional equation three times successively:</p><p>• f(a + x) = b + [1 - g(x)³]^(1/3), so g(a + x) = [1 - g(x)³]^(1/3)</p><p>• g(a + (a + x)) = g(2a + x) = [1 - g(a + x)³]^(1/3) = [1 - (1 - g(x)³)]^(1/3) = [g(x)³]^(1/3) = g(x)</p><p>• g(a + (2a + x)) = g(3a + x) = [1 - g(2a + x)³]^(1/3) = [1 - g(x)³]^(1/3) = g(a + x)</p></p><p><strong>Step 5:</strong> From Step 4, we have g(2a + x) = g(x), which means:</p><p>f(2a + x) - b = f(x) - b</p><p>Therefore: f(2a + x) = f(x) for all x ∈ ℝ</p><p>This proves that f(x) is periodic with period 2a (where a is a positive constant).</p><p><strong>∴ Answer:</strong> f(x) is periodic with period 2a.</p>
Correct Answer: f(x) is periodic