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Calculus 1

Consider the following.
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.)

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  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

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  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

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  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

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

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

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

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

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