When Absolute Values Get Tricky: Differentiability of |f(x)|
Master the critical conditions where absolute value functions lose differentiability. Essential for JEE calculus problems.
Why |f(x)| Differentiability Matters in JEE
Absolute value functions create sharp corners where differentiability fails. Understanding these critical points is essential because:
- JEE frequently tests non-differentiable points in MCQs
- Composite functions with absolute values appear in 2-3 questions per paper
- Graphical interpretation questions rely on this concept
- Continuity vs Differentiability distinction is crucial
The Fundamental Rule: When |f(x)| is NOT Differentiable
$|f(x)|$ is non-differentiable at points where $f(x) = 0$ and $f'(x) \neq 0$
Key Insight:
At points where $f(x) = 0$, the graph of $|f(x)|$ has a sharp corner (cusp) unless the function is flat at that point.
Mathematical Proof:
Consider the left and right derivatives at $x = c$ where $f(c) = 0$:
Left derivative: $\lim_{h \to 0^-} \frac{|f(c+h)| - |f(c)|}{h} = \lim_{h \to 0^-} \frac{|f(c+h)|}{h}$
Right derivative: $\lim_{h \to 0^+} \frac{|f(c+h)| - |f(c)|}{h} = \lim_{h \to 0^+} \frac{|f(c+h)|}{h}$
If $f'(c) \neq 0$, these limits have opposite signs ⇒ Non-differentiable
Problem 1: Basic Absolute Value
Check differentiability of $f(x) = |x-2|$ at $x = 2$
Solution:
Step 1: Identify where inner function equals zero: $x-2 = 0$ at $x=2$
Step 2: Check derivative of inner function: $\frac{d}{dx}(x-2) = 1 \neq 0$
Step 3: Apply the rule: Since $f(2) = 0$ and $f'(2) \neq 0$, $|x-2|$ is non-differentiable at $x=2$
Step 4: Graphical interpretation: V-shaped graph with sharp corner at $(2,0)$
Problem 2: Composite Absolute Function
Find all points where $f(x) = |x^2 - 4|$ is non-differentiable
Solution Approach:
Step 1: Find where inner function equals zero: $x^2 - 4 = 0 \Rightarrow x = \pm 2$
Step 2: Check derivative of inner function: $\frac{d}{dx}(x^2-4) = 2x$
Step 3: Evaluate at critical points:
• At $x = 2$: $f'(2) = 4 \neq 0$ ⇒ Non-differentiable
• At $x = -2$: $f'(-2) = -4 \neq 0$ ⇒ Non-differentiable
Step 4: Non-differentiable points: $x = -2, 2$
The Exception: When |f(x)| CAN Be Differentiable
$|f(x)|$ is differentiable at points where $f(x) = 0$ if $f'(x) = 0$ at those points
Important Exception:
When the function has a horizontal tangent at the zero point, the absolute value function remains smooth.
Example: $f(x) = |x^3|$ at $x = 0$
Step 1: Inner function zero: $x^3 = 0$ at $x=0$
Step 2: Derivative of inner function: $3x^2 = 0$ at $x=0$
Step 3: Since $f(0) = 0$ and $f'(0) = 0$, $|x^3|$ is differentiable at $x=0$
Step 4: In fact, $|x^3| = x^2|x|$, and derivative at 0 is 0
🚀 Quick Solving Strategy
For |f(x)| Differentiability:
- Find all x where f(x) = 0
- Calculate f'(x) at those points
- If f'(x) ≠ 0 → Non-differentiable
- If f'(x) = 0 → Check carefully
Common Pitfalls:
- Don't confuse continuity with differentiability
- |f(x)| is always continuous if f(x) is continuous
- Watch for multiple roots in polynomial functions
- Check both sides of the point
Advanced Problems 4-6 Available in Full Version
Includes trigonometric absolute values, piecewise functions, and JEE Advanced level problems
📝 Quick Self-Test
Test your understanding with these problems:
1. Is $f(x) = |x^2 - 5x + 6|$ differentiable at $x = 2$ and $x = 3$?
2. Find where $f(x) = ||x| - 1|$ is non-differentiable
3. Check differentiability of $f(x) = |\sin x|$ at $x = n\pi$
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