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I want to find the points of discontinuity of the following function:

$$f(x)=\log\left|{\frac{x+2}{x+3}}\right|$$

This is defined for $x\neq-2$ and $x\neq-3$. I proceed to find the first derivative:

$$f'(x)=\frac{1}{x^2+5x+6}$$

This is as well defined for $x\neq-2$ and $x\neq-3$. Since $f(x)$ is not defined for these points either, they should not be critical points. Therefore there should be no critical points. However, my textbook says "$x=-2$, $x=-3$ are points of discontinuity for $f$". Any hints on what I'm doing wrong?

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    $\begingroup$ It's a matter of definitions/conventions. I would not say the functions is discontinuous in $x=-2$ or $x=-3$ because the function doesn't even exist there (the points are not in the function's domain), but perhaps your textbook uses another convention. $\endgroup$
    – StackTD
    Aug 29, 2018 at 15:12
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    $\begingroup$ critical points are not the same as points of discontinuity. There is no need to calculate the derivative of a function in this case. $\endgroup$
    – 5xum
    Aug 29, 2018 at 15:16
  • $\begingroup$ @5xum thanks, points of discontinuity are "removeable" whereas critical points aren't? $\endgroup$
    – Cesare
    Aug 29, 2018 at 15:22
  • $\begingroup$ @Cesare Critical points are minimums and maximums. For example, $0$ is the critical point of $f(x)=x^2$. There is nothing "discontinuous" at $x=0$ for that function. $\endgroup$
    – 5xum
    Aug 29, 2018 at 15:24
  • $\begingroup$ @Cesare On the other hand, for $f'(x)=\sqrt{|x|}$, the derivative at $x=0$ is not defined, but $x=0$ is still a critical point. $\endgroup$
    – 5xum
    Aug 29, 2018 at 15:27

1 Answer 1

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I think I understand it now. A point of discontinuity is when the function is not continous at that point. It is not necessary to derive the function to determine the point of discontinuity. Simply find the domain and the points of discontinuity will be the extremes of the domain. Instead, a critical point refers to minimums and maximums (the function should be derived).

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