Sets, Relations & Functions
Range of a function
Grade 11
Question:
<p>The range of the function \(f(x) = x^2 + \dfrac{1}{x^2 + 1}\) is:</p>
<p>(a) \([1, \infty)\)</p>
<p>(b) \([2, \infty)\)</p>
<p>(c) \(\left[\dfrac{3}{2}, \infty\right)\)</p>
<p>(d) \([5, \infty)\)</p>
Step-by-Step Solution
Key Concept: Decompose the function into two parts: f(x) = x² + 1/(x²+1). Since x² ≥ 0 and 0 < 1/(x²+1) ≤ 1, find the minimum by analyzing the sum of these monotonic components.
<p><strong>Step 1:</strong> Let f(x) = x² + 1/(x²+1). Rewrite the second term: 1/(x²+1) = 1 - x²/(x²+1).</p><p><strong>Step 2:</strong> Observe that x² ≥ 0 (always), and 0 < 1/(x²+1) ≤ 1 (since x²+1 ≥ 1).</p><p><strong>Step 3:</strong> Find the minimum: At x = 0, f(0) = 0 + 1/(0+1) = 1.</p><p><strong>Step 4:</strong> As x → ±∞, x² → ∞ while 1/(x²+1) → 0, so f(x) → ∞.</p><p><strong>Step 5:</strong> To verify 1 is minimum, take derivative: f'(x) = 2x - 2x/(x²+1)² = 2x[1 - 1/(x²+1)²]. Setting f'(x) = 0 gives x = 0 (the only critical point). Since f''(0) > 0, x = 0 is a minimum.</p><p><strong>Step 6:</strong> f is continuous, has minimum value 1 at x = 0, and approaches ∞ as |x| → ∞.</p><p>∴ <strong>Range = [1, ∞)</strong></p>
Correct Answer: A