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

• Mar 9th 2009, 01:29 PM
fattydq
[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.) http://www.webassign.net/www32/symIm...b42e8916c4.gif

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?
• Mar 9th 2009, 07:54 PM
mr fantastic
Quote:

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.) http://www.webassign.net/www32/symIm...b42e8916c4.gif

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.