Consider the following. B(x) = 3x^(2/3) x (a) Find the interval of increase. (En

Question

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.)

Explanation / Answer

(a)   B(x) = 3x^(2/3)-x
       B'(x) = 3(2/3) x^(-1/3) - 1
       B'(x) = 2/x^(1/3) -1 = 0
        2/x^(1/3) = 1
        x^(1/3) = 2
         x = 8

Consider the intervals (0, 8), (8,)

choose any one point from each of the intervals.
if B'(x) < 0 , B(x) is decreasing on that interval
if B'(x) > 0 , B(x) is increasing on that interval

B'(x) = 2/x^(1/3) -1
(0, 8) : choose x=2; B'(2) = .5874 > 0, B is increasing on (0, 8)
(8,) : choose x=10; B'(10) = -0.0716 < 0 , B is decreasing on (8,)

(c) B'(x) = 2x^(-1/3) -1
          B''(x) = 2(-1/3) x^(-4/3)
          B''(x) = -2/3x^(4/3) = 0
          No solution ; no concave up or concave down

(b)       Maximum and minimum:
           B'(x) = 0 , the solution x=8
           B''(x) = -2/3x^(4/3)

         At x=8, B''(x) < 0 , so B(x) has a maximum at x=8
          There is no minimum
        The maximum value is B(8) = 3 (8)^(2/3) - 8 = 4   

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