I'm learning about functions of bounded variations and need some help with this problem:
Given $f(x) = x^{1/3}, x \in [0, 1]$ show that $f \in BV[0, 1]$.
My work and thoughts:
We know that given any function $f: [a, b] \rightarrow \mathbb{R}$, differentiable on the interval $[a, b]$, if $|f'(x)| \leq M \ \forall x \in [a, b]$ (i.e. if the derivative of $f$ is bounded on the closed interval) then $f \in BV[a, b]$.
Considering our function $f(x) = x^{1/3}$ we have $f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}}$. The function $f(x)$ is not differentiable at $0$. How can I work around this problem to bound the derivative for all $x \in [0, 1]$?