Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>The set of points where \( f(x) = \dfrac{x}{1+|x|} \) is differentiable, is</p>
<p>\((-\infty, 0) \cup (0, \infty)\)</p>
<p>\((-\infty, -1) \cup (-1, \infty)\)</p>
<p>\((-\infty, \infty)\)</p>
<p>\((0, \infty)\)</p>
Step-by-Step Solution
Key Concept: Differentiability requires checking the derivative at x = 0 using left and right derivatives separately, since |x| creates a potential point of non-differentiability. The function must be continuous and have equal left and right derivatives at x = 0.
<p><strong>Step 1:</strong> Rewrite f(x) by cases: For x ≥ 0: f(x) = x/(1+x); For x < 0: f(x) = x/(1-x)</p><p><strong>Step 2:</strong> Check continuity at x = 0: lim(x→0⁺) x/(1+x) = 0 and lim(x→0⁻) x/(1-x) = 0, and f(0) = 0 ✓</p><p><strong>Step 3:</strong> Compute right derivative: f'(0⁺) = d/dx[x/(1+x)]|ₓ₌₀ = [(1+x) - x]/(1+x)²|ₓ₌₀ = 1/1 = 1</p><p><strong>Step 4:</strong> Compute left derivative: f'(0⁻) = d/dx[x/(1-x)]|ₓ₌₀ = [(1-x) + x]/(1-x)²|ₓ₌₀ = 1/1 = 1</p><p><strong>Step 5:</strong> Since left and right derivatives are equal (both = 1) and f is continuous everywhere, f is differentiable at x = 0. For x ≠ 0, the function pieces are polynomial quotients with non-zero denominators, so differentiable there too.</p><p>∴ Answer: C (f is differentiable on ℝ or all real numbers)</p>
Correct Answer: C