# Thread: Average value of a function.

1. ## 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

2. $f(c)=\frac{1}{b-a}\int^b_a {f(x)dx}$ when $c\in[a,b]$

You looking for b when f(c)=3.

3. For $F(x) = \int f(x)~dx$ find $F(b) = 3$

4. 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....

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

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

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

Now use the factor thm to find the roots. You know this theorem?

6. um... I don't think I do. could u elaborate more? Thanks

7. You need to colve this cubic $2b+3b^2-b^3 - 3=0$

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 $P(x)$ if $P(a)=0$ then x-a is a factor of $P(x)$

Example to follow is here. The Fundamental Theorem of Algebra

8. Mr. pickslides I convinced you aware that the roots not so "nice"...

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