Find all numbers b such that the average value of f(x)=2+6x−3x^2 on the interval [0,b] is equal to 3. Give your answer as a list of numbers, e.g., {1,2,3}.

To find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 3, we need to use the formula for average value of a function on an interval.

The average value of a function f(x) on the interval [a, b] is given by the following formula:

Average value = 1/(b - a) * ∫[a,b] f(x) dx

In this case, f(x) = 2 + 6x − 3x^2, and we want the average value to be 3 over the interval [0, b]. Therefore, we have:

3 = 1/(b - 0) * ∫[0,b] (2 + 6x − 3x^2) dx

Now, to find the numbers b, we need to solve this equation for b.

First, let's integrate the function:
Integral of (2 + 6x − 3x^2) dx = 2x + 3x^2 - x^3

So our equation becomes:

3 = 1/b * (2b + 3b^2 - b^3)

Now, let's simplify further:

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

Rearranging terms, we get:

0 = b^3 - 3b^2 - b + 6

To solve this cubic equation, we can use numerical methods or algebraic methods such as factoring or synthetic division. However, in this case, let's use a numerical method called the "guess and check" method.

We can start by trying different values of b and evaluating the equation to see if it equals 0. Let's start with integers.

Trying b = 1:
1^3 - 3(1)^2 - 1 + 6 = 3 - 3 - 1 + 6 = 5, not equal to 0.

Trying b = 2:
2^3 - 3(2)^2 - 2 + 6 = 8 - 12 - 2 + 6 = 0

Therefore, b = 2 is a solution to the equation.

Hence, the list of numbers b such that the average value of f(x) = 2 + 6x − 3x^2 on the interval [0, b] is equal to 3 is {2}.

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