Limits, Continuity & Differentiability
Differentiability of Min/Max Functions
Grade 12
Question:
<p>Let \(h(x) = \min\{x, x^2\}\) for every real number \(x\). Then,</p>
<p>(a) \(h\) is not continuous for all \(x\)</p>
<p>(b) \(h\) is differentiable for all \(x\)</p>
<p>(c) \(h'(x) = 2\) for all \(x > 1\)</p>
<p>(d) \(h\) is not differentiable at two values of \(x\)</p>
Step-by-Step Solution
Key Concept: To analyze h(x) = min{x, x²}, we must first determine where x < x² and where x > x², then examine continuity and differentiability at transition points. A function is differentiable only if the left and right derivatives exist and are equal at that point.
<p><strong>Step 1: Determine where each piece dominates</strong><br>Compare x and x²:<br>• x = x² when x² - x = 0 ⟹ x(x-1) = 0 ⟹ x = 0 or x = 1<br>• For x < 0: x < x² (since x² > 0 > x)<br>• For 0 < x < 1: x > x² (since x² < x)<br>• For x > 1: x < x² (since x² > x)<br>Therefore: h(x) = {x² if x ≤ 0, x if 0 < x < 1, x² if x ≥ 1}</p><p><strong>Step 2: Check continuity</strong><br>At x = 0: lim(x→0⁻) h(x) = 0, lim(x→0⁺) h(x) = 0, h(0) = 0 ✓ Continuous<br>At x = 1: lim(x→1⁻) h(x) = 1, lim(x→1⁺) h(x) = 1, h(1) = 1 ✓ Continuous<br>h is continuous for all x ∈ ℝ</p><p><strong>Step 3: Check differentiability at x = 0</strong><br>Left derivative: lim(h→0⁻) [h(h) - h(0)]/h = lim(h→0⁻) h²/h = lim(h→0⁻) h = 0<br>Right derivative: lim(h→0⁺) [h(h) - h(0)]/h = lim(h→0⁺) h/h = 1<br>Left derivative ≠ Right derivative ⟹ h is NOT differentiable at x = 0</p><p><strong>Step 4: Check differentiability at x = 1</strong><br>Left derivative: lim(h→0) [h(1+h) - h(1)]/h = lim(h→0⁻) [(1+h) - 1]/h = 1<br>Right derivative: lim(h→0) [h(1+h) - h(1)]/h = lim(h→0⁺) [(1+h)² - 1]/h = lim(h→0⁺) [2h + h²]/h = 2<br>Left derivative ≠ Right derivative ⟹ h is NOT differentiable at x = 1</p><p><strong>Step 5: Check other statements</strong><br>• (a) False: h is continuous for all x ✓<br>• (b) False: h is not differentiable at x = 0 and x = 1<br>• (c) False: h'(x) = 2x for x > 1, not constant 2<br>• (d) True: h is not differentiable at exactly x = 0 and x = 1 (two values)</p><p><strong>∴ Answer: d</strong></p>
Correct Answer: d