Limits, Continuity & Differentiability
Differentiability of functions
Grade None
Question:
<p>It is given that <br>
\(f(x) = \text{Min}\{x+1, |x|+1\}\). <br>
Then which of the following is true?</p>
<p>\(f(x) \geq 1\) for all \(x \in R\)</p>
<p>\(f(x)\) is not differentiable at one point</p>
<p>\(f(x)\) is differentiable everywhere</p>
<p>\(f(x)\) is not differentiable at two points</p>
Step-by-Step Solution
Key Concept: The function f(x) is the minimum of two expressions at each point: you must evaluate both y = x+1 and y = |x|+1, then take the lower value. The critical behavior occurs where these curves intersect and at the cusp of the absolute value.
<p><strong>Step 1:</strong> Identify the two component functions: g(x) = x+1 and h(x) = |x|+1.</p><p><strong>Step 2:</strong> Find intersection points. For x ≥ 0: x+1 = x+1 (always equal). For x < 0: x+1 = -x+1, giving x = 0.</p><p><strong>Step 3:</strong> Analyze which function is minimum in each region:</p><ul><li>For x ≥ 0: Both equal x+1, so f(x) = x+1</li><li>For x < 0: Compare x+1 vs -x+1. Since -x+1 > x+1 when x < 0, we have f(x) = x+1</li></ul><p><strong>Step 4:</strong> Therefore f(x) = x+1 for all x ∈ ℝ.</p><p><strong>Step 5:</strong> This linear function is continuous everywhere and differentiable everywhere with f'(x) = 1.</p><p>∴ Answer: C</p>
Correct Answer: C