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The derivative of the function $f$ is,

$$ f'(x)=\frac{x^2(x-1)}{x+2},~ x\ne-2. $$

The obvious critical points are $x=0$ and $x=1$. However, since $f'(-2)$ is not defined, can $-2$ be a critical point?

I have read that for $c$ to be a critical number/point, $f(c)$ must be well-defined, and $f'(c)$ must be $0$ or undefined. However, we are not given $f$. Assuming we have no way to integrate this function to check whether $f(-2)$ is defined or graph the given derivative, is it possible to consider $-2$ to be a critical number?

If not, how can we justify using $-2$ to split the domain of $f$ into sub-intervals along with the other critical numbers? The sub-intervals in which I test the trend of the function (increasing/decreasing) are formed using the critical numbers and $-2$, as the textbook dictates. So my intervals would be $(-\infty,-2)$, $(-2,0)$, $(0,1)$ and $(1,\infty)$.

PS: This textbook uses the terms "critical numbers" and "critical points" interchangeably.

rayank97
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  • If the domain of a function like this is unspecified, it is not unreasonable to assume the function is $f: \mathbb R \mapsto \mathbb R$. – whoisit Oct 03 '22 at 15:27
  • At any rate, $x=-2$ is clearly an important boundary point when understanding the behaviour of $f(x)$ on various intervals, irrespective of whether $f(-2)$ is defined. – Greg Martin Oct 03 '22 at 15:57
  • @GregMartin What would lead one to say that $x=-2$ is an important boundary point without looking at the graph of the derivative? – rayank97 Oct 03 '22 at 17:52
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    The derivative is undefined and hence not continuous there; that alone is important (when we're finding intervals on which the function is increasing/decreasing). – Greg Martin Oct 04 '22 at 00:10

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