Functions like \(f^{'}(x) = \frac{x^2*(x-1)}{x+2}; x \neq -2\) has critical pts of \(1\) and \(0\), but then to find the increasing and decreasing values of the function, \(-2\) is added in also.
While functions like \(f^{'}(x) = (x^{\frac{-1}{3}})(x+2)\) has critical pts of \(-2\) *and* \(0\), while \(0^{\frac{-1}{3}}\) is undefined.
I hope this all makes sense!
When are the undefined points of a function's derivative critical points?
This is probably an easy question for you guys, but the not-so-bright me cannot figure it out.

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TopNewestcritical points are the points for which the value of the function is either zero or it is not defined .. say the function approaches to infinity for the particular point ... the same is the case when u say 0^(-1/3) is undefined since f'(x) approaches to infinity for x=0 and for x=-2 its value is zero .. hope that makes sense.. :) – Ramesh Goenka · 2 years, 8 months ago

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