Question

need some help setting the derivative of the function below equal to zero to complete a max/min problem

Answer 1

I got as far as f'(x)=(4/3)x^(1/3) - (16/3)x^(-2/3) but I don't know how to find the x criticals from that

Answer 2

Alright, let me just type up that derivative so we can see it better :P

Answer 3

thanks sorry i'm not so savvy :)

Answer 4

thanks, the only thing i dont see is where the 12x came from
shouldnt I only multiply the equation by the left by x^(2/3) because it already has a 3 in the numerator?

Answer 5

Yeah, you're correct, my mistake. I got it stuck in my head that the common denominator was the 3x^(2/3) and accidentally multiplied the 3up, too x_x

Answer 6

I delete that before I make myself look like more of an idiot than what I am. Oh well, process is correct x_x

Answer 7

thanks :)
is the next step to find the maxs and mins to plug the x critical (16) and the intervals into the original function?

Answer 8

Yes, whatever you find your critical point to be you would plug into the original as well as the end points of your interval :3

Answer 9

thank you
I got
f(-1)=17 (max)
f(16)=0
f(8)=-16 (min)
could you please varify? I think the maxs and mins are supposed to have decimals for this problem . . .

Answer 10

I wasn't sure what your interval was xD

Answer 11

oh the interval is [-1,8]

Answer 12

So wait, you got 16 for your critical point?

Answer 13

yeah, but I also don't know how I got that because its not within the interval

Answer 14

Right. So if you solve for x in the numerator, you get x = 4

Answer 15

yepppp
XO
haha what a silly mistake thank you so much for your help

Answer 16

Lol, right xD So -19.04 or so for the critical point and I didnt check the intervals xD