<p>The given function is \(f(x) = (x+1)^{1/3} - (x-1)^{1/2}\); \(x \in [0,1]\). What is the greatest value of \(f(x)\) on \([0,1]\)?</p>
Step-by-Step Solution
Key Concept: For functions on closed intervals, the maximum occurs either at critical points (where f'(x)=0) or at boundary endpoints. You must evaluate f at all such points and compare.
<p><strong>Step 1:</strong> Find the derivative of f(x) = (x+1)^(1/3) - (x-1)^(1/2)</p><p>f'(x) = (1/3)(x+1)^(-2/3) - (1/2)(x-1)^(-1/2)</p><p><strong>Step 2:</strong> Set f'(x) = 0 to find critical points:</p><p>(1/3)(x+1)^(-2/3) = (1/2)(x-1)^(-1/2)</p><p>Rearranging: 2(x-1)^(-1/2) = 3(x+1)^(-2/3)</p><p>This equation is transcendental and difficult to solve analytically. Analyze the sign of f'(x) on (0,1): as x increases, (x+1)^(-2/3) decreases slowly while (x-1)^(-1/2) decreases rapidly (becomes undefined at x=1). Thus f'(x) > 0 throughout [0,1), meaning f is strictly increasing on [0,1].</p><p><strong>Step 3:</strong> Since f is increasing on [0,1], the maximum occurs at the right endpoint x = 1.</p><p>f(1) = (1+1)^(1/3) - (1-1)^(1/2) = 2^(1/3) - 0 = 2^(1/3) = ∛2</p><p><strong>Step 4:</strong> Verify by checking the left endpoint: f(0) = 1^(1/3) - 1 = 0, which is less than ∛2.</p><p>∴ Answer: A (Maximum value = ∛2)</p>
Correct Answer: A