# Average value of a function.

• Jul 28th 2010, 04:38 PM
softballchick
Average value of a function.
Find the number b such that the average value of f(x)=2+6x-3x^2 on the intervak [0, b] is equal to 3.

so I did the integal from 0 to b of 2+6x-3x^2 which is 2x+3x^2-x^3

now what? Do i set it equal to 3 and find out what b is?

Thanks
• Jul 28th 2010, 04:42 PM
Also sprach Zarathustra
$\displaystyle f(c)=\frac{1}{b-a}\int^b_a {f(x)dx}$ when $\displaystyle c\in[a,b]$

You looking for b when f(c)=3.
• Jul 28th 2010, 04:45 PM
pickslides
For $\displaystyle F(x) = \int f(x)~dx$ find $\displaystyle F(b) = 3$
• Jul 28th 2010, 05:15 PM
softballchick
right, so I set 2x+3x^2-x^3 = 3 right?

= x(2+3x-x^2)
=x(x -1) (x-2)

ahhh i'm confused....
• Jul 28th 2010, 05:20 PM
pickslides
Quote:

Originally Posted by softballchick
right, so I set 2x+3x^2-x^3 = 3 right?

Yep, then

$\displaystyle 2b+3b^2-b^3 = 3$

$\displaystyle 2b+3b^2-b^3 - 3=0$

Now use the factor thm to find the roots. You know this theorem?
• Jul 28th 2010, 05:28 PM
softballchick
um... I don't think I do. could u elaborate more? Thanks
• Jul 28th 2010, 05:44 PM
pickslides
You need to colve this cubic $\displaystyle 2b+3b^2-b^3 - 3=0$

To do that you should employ the factor theorem. It's really not something that can be explained very well in a few lines but here goes.

The Factor Theorem: For the polynomial $\displaystyle P(x)$ if $\displaystyle P(a)=0$ then x-a is a factor of $\displaystyle P(x)$

Example to follow is here. The Fundamental Theorem of Algebra
• Jul 28th 2010, 05:58 PM
Also sprach Zarathustra
Mr. pickslides I convinced you aware that the roots not so "nice"...
• Jul 28th 2010, 06:06 PM
pickslides
Quote:

Originally Posted by Also sprach Zarathustra
Mr. pickslides I convinced you aware that the roots not so "nice"...

I did not look that closely!