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Question

Find the numbers b such that the average value of f(x) = 7 + 6x - 3x^2 on the interval [0, b] is equal to 8.

b = (smaller value)
b = (larger value)

Answer 1

ave value [ ∫ 7 + 6x - 3x^2 dx ]/b = 8 .... from 0 to b.
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7x + 3x^2 - x^3 from 0 to b = 8b
7b + 3b^2 - b^3 = 8b
b^3 -3b^2 + b = 0
b(b^2 -3b + 1) =0
b^2 - 3b + 9/4 = 5/4
b - 3/2 = ±√5 / 2

b = [3-√5]/2

b = [3+√5]/2

Answer 2

if 0.b is equal to 8 then

b= larger value if u count it from 1-10.
8 is closer to the number 10

Answer 3

Average value = (f(b)-f(0))/(b-0)
= (7+6b -3b^2 -7)/b = 8
6b-3b^2 =8b
-2b -3b^2= 0
-b(2+3b) = 0
b= -2/3

Answer 4

integrating f(x) wrt x in the interval [0,b].... we get 7b + 3b^2 - b^3

for avg, integral f(x) dx/ integraldx
we get (7b+ 3b^2 - b^3)/b = 8
or 7b+ 3b^2 - b^3 = 8b
or b^3 - 3b^2 + b = 0
or b(b^2 - 3b +) = 0
hence we get three roots,
b=0
b= [3+sqrt(5)]/2
b= [3-sqrt(5)]/2

Answer 5

idk