Consider the following.

Question

B(x) = 3x^(2/3) − x

(a) Find the interval of increase.(Enter your answer using interval notation.)
Find the interval of decrease. (Enter your answer using interval notation.)

(b) Find the local maximum value(s).(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
Find the local minimum value(s).(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

(c) Find the interval where the graph is concave downward. (Enter your answer using interval notation.)

Answer 1

why is this causing difficulties?

B'(x) = 2/∛x - 1
B'(x) = 0 at x=8

as you will recall,
B is increasing where B' > 0
B is concave up where B" > 0

Answer 2

Please don't assist me anymore cuz you always seem to make me feel bad. I work hard. Every now and then I get a few problems that I'm unsure of if I'm doing correctly. Geez

Answer 3

Sorry. I might be less grumpy if you actually showed some of that hard work.

Given the info I provided above,

(a) B is increasing where B' > 0, so
2/∛ - 1 > 0
2/∛x > 1
2 > ∛x
8 > x
So, B is increasing on (0,8)
B is decreasing on (-∞,0)U(8,+∞)

(b) B'= 0 at x=8, so that's a max or a min.
B" = -2/(3x^(5/3))
B"(8) < 0, so B(8) is a max.
There are no minima.

(c) B" is always negative, so B is always concave down: (-∞,+∞)

The graph is at

http://www.wolframalpha.com/input/?i=3%28x^2%29^%281%2F3%29-x

Answer 4

There is a local minima actually, and 8 is not the local max

Answer 5

AcTuAlLy if you BoThEreD to do WERK (sis) then you might know der!v@t!ve$

Answer 6

Actually there is a local minimum at the point (0,0)

Answer 7

The minimum value is at (0, 0) and
The maximum value is at (8, 4)