Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>Let \( f: R \to R \) be a function defined by \( f(x) = \min\{x+1,\ |x|+1\} \). 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 \( x = 1 \).</p>
<p>\( f(x) \) is differentiable everywhere.</p>
<p>\( f(x) \) is not differentiable at \( x = 0 \).</p>
Step-by-Step Solution
Key Concept: Analyze the piecewise definition of f(x) by comparing x+1 and |x|+1 in different regions, then check continuity and differentiability at transition points where the minimum switches.
<p><strong>Step 1:</strong> Determine f(x) by comparing x+1 and |x|+1.</p><p>For x ≥ 0: |x|+1 = x+1, so both are equal. Thus f(x) = x+1.</p><p>For x < 0: |x|+1 = -x+1. Compare: x+1 vs -x+1 gives x+1 < -x+1 ⟹ x < 0 (true). So f(x) = x+1.</p><p>Therefore: <strong>f(x) = x+1 for all x ∈ ℝ</strong></p><p><strong>Step 2:</strong> Verify continuity and differentiability.</p><p>Since f(x) = x+1 is a linear function on all of ℝ:</p><p>• f is continuous everywhere on ℝ</p><p>• f'(x) = 1 everywhere on ℝ</p><p>• f is differentiable everywhere on ℝ</p><p>∴ Answer: C (f is continuous and differentiable everywhere on ℝ)</p>
Correct Answer: C