Question about 2nd-derivative test for finding relative max/min (posting below)

Question

anonymous · Answer

I need to find the relative max/min values for this function:
$$g(x) = x + 2/\sqrt{x}$$

I found the 1st derivative which is g'(x) = 1 + (-x^-3/2)
I found the critical numbers which are 0 and 1

Then the 2nd derivative, which is g''(x) = (3/2)x^(-5/2)
g''(0) = 0
g''(1) > 0
(1,f(1)) is a relative minimum

But what does it mean when g''(0) = 0?

anonymous · Answer

Wait ya stupid mistake. 0 is not a critical number

anonymous · Answer

sorry for my silly mistake:p

aum · Answer

Well, you realized 0 is not a critical number.
Also, g''(0) is not zero but undefined.

imstuck · Answer

Not to butt in, but how is 0 not a critical number? When you set the second derivative = to 0, then x = 0, doesn't it? @aum

aum · Answer

x = 0 is not in the domain of g(x).
g(x) is only defined for x > 0

$\Large g''(x) = \frac 32 \frac{1}{x^{5/2}}$
So g''(0) is undefined.

aum · Answer

We say that x = c is a critical point of the function f(x), 
IF f(c) exists AND either f'(c) = 0 OR f'(c) does not exist.
Note that we require that  f(c) exists in order for x = c  to actually be a critical point.  This is an important, and often overlooked, point.

Here g(0) does not exist and therefore x = 0 is not a critical point.