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?
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.
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.