The range of the function f(x) = <span class="math-inline">\(e^{x}+e^{x}\)</span>, is -
Step-by-Step Solution
Key Concept: Use the AM-GM inequality to find the minimum value of e^x + e^(-x). Since both e^x and e^(-x) are always positive, their sum achieves a minimum when they are equal.
<p><strong>Step 1:</strong> Identify the function: f(x) = e^x + e^(-x) (interpreting the given notation correctly).</p><p><strong>Step 2:</strong> Apply the AM-GM inequality. For positive numbers a and b: (a + b)/2 ≥ √(ab), which gives a + b ≥ 2√(ab).</p><p><strong>Step 3:</strong> Let a = e^x and b = e^(-x), both always positive for all real x. Then: e^x + e^(-x) ≥ 2√(e^x · e^(-x)) = 2√(e^(x-x)) = 2√(e^0) = 2√1 = 2.</p><p><strong>Step 4:</strong> Verify when equality holds. Equality in AM-GM occurs when a = b, so e^x = e^(-x), which gives e^(2x) = 1, so x = 0. At x = 0: f(0) = e^0 + e^0 = 1 + 1 = 2. ✓</p><p><strong>Step 5:</strong> As x → ±∞, at least one of e^x or e^(-x) → ∞, so f(x) → ∞. The minimum value is 2, achieved at x = 0, and the function has no upper bound.</p><p><strong>∴ Answer:</strong> C (f(x) ≥ 2)</p>
Correct Answer: C