Do critical numbers exist at an undefined point?

[jeonw]

Do critical numbers exist where the denominator equals zero?
For example my calculator says -2 and 2 are undefined when the function is 1/(x^2-4).
However my calculator says 0 exist when the function is 2/x^(2/3).

Can someone clarify when critical numbers exist and don't exist.

 

[mmm4444bot]

my calculator says -2 and 2 are undefined

my calculator says 0 exist

I have to guess what you mean because your English is not correct.

Here is a definition for "critical number":

A critical number for a function is any number in the function's domain that causes the function's first derivative to equal zero OR to be undefined.

\(\displaystyle f(x) \;=\; \frac{1}{x^2 \;-\; 4}\)

\(\displaystyle f'(x) = -\frac{2}{(x^2 \;-\; 4)^2}\)

f`(x) is not defined for x = -2 or x = 2; however, -2 and 2 are not in the domain of function f.

Therefore, function f has no critical numbers.

\(\displaystyle g(x) \;=\; \frac{2}{x^{2/3}}\)

\(\displaystyle g'(x) = -\frac{4}{3} \cdot \frac{x}{x^{8/3}}\)

g`(x) is not defined for x = 0; however, 0 is not in the domain of function g.

Therefore, function g has no critical numbers.

.

Do critical numbers exist where the denominator equals zero?

If you're talking about the denominator of the first derivative, then the answer is yes IF the original function exists at those points.

ANY critical number must be in the domain of the original function.