Question

Check the differentiability of the function f(x) = |x| at x = 0.

Hint:

To check differentiability at $x = 0$, we examine the left-hand and right-hand derivatives.

Step 1: Consider the function $f(x) = |x|$

• For $x > 0$, $f(x) = x$ ⟹ $f'(x) = \mathbf{1}$
• For $x < 0$, $f(x) = -x$ ⟹ $f'(x) = \mathbf{-1}$

Step 2: Evaluate the one-sided derivatives at $x = 0$

• Left-hand derivative: $f'_-(0) = \mathbf{-1}$
• Right-hand derivative: $f'_+(0) = \mathbf{1}$

Conclusion:

Since $f'_-(0) \ne f'_+(0)$, the function $f(x) = |x|$ is not differentiable at $x = 0$.

Answer 1

Given: $f(x) = |x|$

Step 1: Define $f(x)$ piecewise

$f(x) = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Step 2: Check continuity at $x = 0$

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$
$f(0) = |0| = 0$
$\Rightarrow$ Function is continuous at $x = 0$

Step 3: Check differentiability at $x = 0$

Left-hand derivative:

$f'_-(0) = \lim_{h \to 0^-} \dfrac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \dfrac{-h - 0}{h} = \lim_{h \to 0^-} (-1) = \mathbf{-1}$

Right-hand derivative:

$f'_+(0) = \lim_{h \to 0^+} \dfrac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \dfrac{h - 0}{h} = \lim_{h \to 0^+} (1) = \mathbf{1}$

Conclusion:

Since $f'_-(0) \ne f'_+(0)$, the function $f(x) = |x|$ is not differentiable at $x = 0$.