Question

Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=(x-3)^{2 / 3} $$

Short answer

The function \(y=(x-3)^{2 / 3}\) is differentiable at all \(x\)-values except at \(x=3\).

Step-by-step solution

Calculate the derivative of the function

To start with, it is needed to find the derivative of the function \(y=(x-3)^{2/3}\). By using the chain rule (the derivative of a composite function), the derivative (\(y' \) or \(\frac{dy}{dx}\)) of the function can be found as follows: \(y'=\frac{2}{3}(x-3)^{-1/3}\cdot 1\). Which simplifies to: \(y'=\frac{2}{3\sqrt[3]{(x-3)}}\).

Analyze the derivative function

Upon scrutinizing the derivative, it is observed that it is undefined when the denominator is zero, i.e., \(\sqrt[3]{(x-3)}=0\). Solving this equation gives the root as \(x=3\). Therefore, the function \(y=(x-3)^{2 / 3}\) is not differentiable at \(x=3\).

Identify x-values for differentiability

Having found the point of non-differentiability, one can mention that the function \(y=(x-3)^{2/3}\) is differentiable at all other \(x\)-values in its domain, i.e., for all \(x\neq3\).