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 rewritten as: \[ f(x) = \begin{cases} x^3, & \text{if } x \geq 0,
(-x)^3 = -x^3, & \text{if } x > 0. \end{cases} \] 1. Continuity: At \( x = 0 \), the left-hand and right-hand limits are: \[ \lim_{x \to 0^-} f(x) = (-x)^3 = 0, \quad \lim_{x \to 0^+} f(x) = x^3 = 0. \] Since both limits and \( f(0) = 0 \), the function is continuous. 2. Differentiability: Differentiate \( f(x) \) for \( x > 0 \) and \( x < 0 \): \[ f'(x) = \begin{cases} 3x^2, & \text{if } x > 0,
3(-x)^2 = 3x^2, & \text{if } x < 0. \end{cases} \] At \( x = 0 \), compute the left-hand and right-hand derivatives: \[ f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{-x^3}{x} = \lim_{x \to 0^-} -x^2 = 0, \] \[ f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{x^3}{x} = \lim_{x \to 0^+} x^2 = 0. \] Since \( f'(0^-) = f'(0^+) = 0 \), the derivative exists at \( x = 0 \). Therefore, \( f(x) = |x|^3 \) is differentiable at all points of its domain.