Question

The set of points where the function \( f(x) = |x| \) is differentiable, is

• A. \( (-\infty, \infty) \)
• B. \( (-\infty, 0) \cup (0, \infty) \)
• C. \( [0, \infty) \)
• D. \( [0, \infty) \cup (-\infty, 0) \)

Hint:

The absolute value function is not differentiable at \( x = 0 \) because the left-hand and right-hand derivatives do not match.

Answer 1

The Correct Option is A

Step 1: Recall the definition of differentiability.
A function is differentiable at a point if its derivative exists at that point. The absolute value function \( f(x) = |x| \) is piecewise-defined: \[ f(x) = \begin{cases} x & \text{for } x \geq 0 -x & \text{for } x<0 \end{cases} \]

Step 2: Differentiate the function for \( x \geq 0 \) and \( x<0 \).
For \( x \geq 0 \), \( f(x) = x \), so the derivative is:
\[ f'(x) = 1 \] For \( x<0 \), \( f(x) = -x \), so the derivative is:
\[ f'(x) = -1 \]

Step 3: Analyze the behavior at \( x = 0 \).
At \( x = 0 \), we must check the left-hand and right-hand derivatives. The left-hand derivative (approaching 0 from the negative side) is:
\[ \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{-x}{x} = -1 \] The right-hand derivative (approaching 0 from the positive side) is: \[ \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{x}{x} = 1 \]

Step 4: Conclusion.
Since the left-hand and right-hand derivatives at \( x = 0 \) are not equal, the function is not differentiable at \( x = 0 \). However, the function is differentiable everywhere else. Therefore, the set of points where \( f(x) = |x| \) is differentiable is \( (-\infty, 0) \cup (0, \infty) \), corresponding to option (A).

Step 5: Verification.
The absolute value function is continuous everywhere, but it is only non-differentiable at \( x = 0 \) because of the discontinuity in the derivative.