# Thread: [SOLVED] Average Value Of a Function question with a twist

1. ## [SOLVED] Average Value Of a Function question with a twist

Find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 11. (Enter solutions in exact forms from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.)

Now I'm assuming I'd do 11=1/x[8x+5x^2-x^3] which simplifies to x^2-5x-3=0, but I don't know how to solve for x from there, especially to get two different solutions (which is what the answer key of the problem asks for). Could someone help me out?

2. Originally Posted by fattydq
Find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 11. (Enter solutions in exact forms from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.)

Now I'm assuming I'd do 11=1/x[8x+5x^2-x^3] which simplifies to x^2-5x-3=0, but I don't know how to solve for x from there, especially to get two different solutions (which is what the answer key of the problem asks for). Could someone help me out?
Solve $\displaystyle \frac{\int_0^b f(x) \, dx}{b} = 11 \Rightarrow -b^2 + 5b + 8 = 11 \Rightarrow b^2 - 5b + 3 = 0$ for $\displaystyle b$. I suggest using the quadratic formula.