Register to reply

Finding the critical numbers

by loadsy
Tags: critical, numbers
Share this thread:
loadsy
#1
Oct28-06, 10:46 PM
P: 57
I have a problem that involves trying to find the critical numbers of a function.

g(x) = x^1/3 - x^-2/3

I would assume in this instance that Dg = [0, +infinity)
g'(x) = 1/3x^-2/3 - (-2/(3(x^-5/3))
g'x = 1/ 3(x^2/3) + 2/ 3(x^5/3)

Now could you say that g'(0) = undefined, so 0E[0, +infinity) and g'(0) d.n.e therefore, 0 is a critical number of g. However, are there are any other critical numbers in this instance? And if so how do you solve for that?
Phys.Org News Partner Science news on Phys.org
Suddenly, the sun is eerily quiet: Where did the sunspots go?
'Moral victories' might spare you from losing again
Mammoth and mastodon behavior was less roam, more stay at home
HallsofIvy
#2
Oct29-06, 09:59 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,285
Yes, certainly, since g' is not defined at x= 0, that is, by definition, a critical number. That is the only value of x where the derivative is not defined but there might be other critical numbers where the derivative is equal to 0. How do you solve for that? Solve the equation 1/(3x^(2/3))+ 2/(3x^(5/3))= 0, of course. Looks pretty straight forward to me.
loadsy
#3
Oct29-06, 12:36 PM
P: 57
Ahhhh I follow you now. So the other critical number after solving 1/(3x^(2/3))+ 2/(3x^(5/3))= 0, I got was x= -2 in this instance. I'm assuming this is another critical number in this question. Therefore since this question is only asking for the "critical numbers", they are then just 0 and -2 I would assume.


Register to reply

Related Discussions
Critical numbers Calculus & Beyond Homework 2
Finding Critical Numbers Calculus & Beyond Homework 7
Finding Critical Numbers Calculus & Beyond Homework 3
Finding Critical numbers Introductory Physics Homework 5