Showing That a Function Has No Derivative

Show that \(f( x) =\sqrt[5]{x+2}\) has no derivative at \(x=-2\).

Solution First we note that \(f(-2) =\sqrt[5]{-2+2}=0.\) Then we investigate the limit. \[ \lim\limits_{x\rightarrow -2}\dfrac{f( x) -f(-2) }{ x-(-2) }=\lim\limits_{x\rightarrow -2}\dfrac{\sqrt[5]{x+2}-0}{x+2 }=\lim\limits_{x\rightarrow -2}\dfrac{\sqrt[5]{x+2}}{x+2}=\lim\limits_{x \rightarrow -2}\dfrac{1}{\sqrt[5]{( x+2) ^{4}}}=\infty \]

Since the limit is infinite, the derivative of \(f\) does not exist at \(-2\).