Register to reply 
Finding the critical numbers 
Share this thread: 
#1
Oct2806, 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? 


#2
Oct2906, 09:59 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

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.



#3
Oct2906, 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 