Find all numbers b such that the average value of f(x)=5+10x-9x^2 on the interval [0, b] is equal to 6

Question

Please help! This is what I've done so far but I know it's not right...

I used the formula from the mean value theorem to get to

6=(1/b) times the integral from 0 to b of (5+10x-9x^2)dx

then I got to the equation 5b+5b^2-3x^3=6b

-b+5b^2-3b^3=0

b=0, (5+/- sqrt13)/2

Answer 1

there are two different mean value theorems, one for derivatives(tangent slope and line slope) and one for integrals (area and area/length is height).

obviously the problem is referencing integals theorem mean value.

the equation should be to find b satisfying requirement:

3b²-5b+1=0,,,,

not 5b + 10b - 9b² - 5 = 6b

because

wanted 6=(1/b)*(5b+5b²-(1/3)b³)

by the way, b₂≈1.43

(1/1.43)*(-3*1.43³+5*1.43²+5*1.43)≈6

b₁≈.23

note f(x)=6 when x₂=1 and x₁≈.12

so b₂>x₂ and b₁>x₁

whew.



Answer 2

The Mean Value Theorem states at least one point on the function has a tangent parallel to the secant through the end points. "Average Value" could mean:

[f(b) - fI0)] / b = 6

f(b) - f(0) = 6b

5b + 10b - 9b² - 5 = 6b