Question

How do you find the maxima and minima for g(x) = x^2 - 2/x

Answer 1

First note that g(x) is not defined when x = 0. So the domain of g(x) excludes x.
0 = 2x + 2/x^2
multiply by x^2
0 = 2x^3 + 2 = 2(x^3 + 1)
x^3 = -1
x = -1.

Answer 2

I meant - So the domain of g(x) excludes 0.

Answer 3

Well, see my note above. x = 0 is excluded from the domain of g(x).

Answer 4

But the original function does not exist at x = 0 and so it is excluded from g'(x) or g''(x).

Answer 5

The function has to be continuous in an interval. g(x) is not continuous at x = 0. So we exclude x = 0.

Answer 6

You should be... critical numbers are where the *derivative* is either zero or undefined.
^_^

Answer 7

If a function itself is not defined at a point, then there is no...point in pondering whether that undefined point could be a max or a min... am I right? :D

Answer 8

So, I guess that about wraps things up? :D

Answer 9

I should think so ^_^

Answer 10

But the question is about finding extrema... is this critical number an extremum at all? ^^

Answer 11

Fair enough ^_^

Answer 12

A positive second derivative implies upward concavity... just saying ;)

Answer 13

Well, positive second derivative means that it's curving upward, more or less

Answer 14

Precisely ^_^
(this is assuming that the first derivative was zero, mind you ;) )

Answer 15

No problem.

Answer 16

It does look like @aum did most of the heavy lifting though.
Just saying... again ;)

Answer 17

no big deal. As long as the OP gets what he was seeking for.

Answer 18

The joy of learning...
bliss :D

Answer 19

I looked up the definition of critical points:

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.

http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx

Answer 20

Here f(0) does not exist. Therefore, x = 0 is not a critical point.