Question

(b) Show that the function \( f(x) = |x|^3 \) is differentiable at all points of its domain.

Hint:

To check differentiability of piecewise functions, confirm continuity and ensure that the left-hand and right-hand derivatives are equal at the joining point.

Answer 1

The function \( f(x) = |x|^3 \) can be expressed as:\[f(x) =\begin{cases}x^3, & \text{if } x \geq 0,
-x^3, & \text{if } x<0.\end{cases}\]1. Continuity: At \( x = 0 \), the one-sided limits are: \[ \lim_{x \to 0^-} f(x) = 0, \quad \lim_{x \to 0^+} f(x) = 0. \] As these limits equal \( f(0) = 0 \), the function is continuous.2. Differentiability: The derivative of \( f(x) \) for \( x eq 0 \) is: \[ f'(x) = \begin{cases} 3x^2, & \text{if } x>0,
3x^2, & \text{if } x<0. \end{cases} \] At \( x = 0 \), the one-sided derivatives are: \[ f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = 0, \] \[ f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = 0. \] Since \( f'(0^-) = f'(0^+) = 0 \), the derivative exists at \( x = 0 \). Consequently, \( f(x) = |x|^3 \) is differentiable for all \( x \).