To find the values of \( b \) such that the average value of the function
\[ f(x) = 3 + 10x - 9x^2 \]
on the interval \([0, b]\) is equal to 4, we first need to calculate the average value of the function on that interval. The average value of a continuous function \( f(x) \) on the interval \([a, b]\) is given by the formula:
\[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) , dx \]
In this case, \( a = 0 \) and \( b = b \), so the formula simplifies to:
\[ \text{Average value} = \frac{1}{b} \int_0^b f(x) , dx \]
We set this equal to 4:
\[ \frac{1}{b} \int_0^b f(x) , dx = 4 \]
This implies:
\[ \int_0^b f(x) , dx = 4b \]
Next, we need to compute the integral \( \int_0^b f(x) , dx \):
\[ \int_0^b f(x) , dx = \int_0^b (3 + 10x - 9x^2) , dx \]
We can compute this integral:
\[ \int_0^b f(x) , dx = \int_0^b 3 , dx + \int_0^b 10x , dx - \int_0^b 9x^2 , dx \]
Calculating each term:
- \(\int_0^b 3 , dx = 3b\)
- \(\int_0^b 10x , dx = [5x^2]_0^b = 5b^2\)
- \(\int_0^b 9x^2 , dx = [3x^3]_0^b = 3b^3\)
Putting it all together:
\[ \int_0^b f(x) , dx = 3b + 5b^2 - 3b^3 \]
Setting this equal to \( 4b \):
\[ 3b + 5b^2 - 3b^3 = 4b \]
Rearranging gives:
\[ -3b^3 + 5b^2 - b = 0 \]
Factoring out \( b \):
\[ b(-3b^2 + 5b - 1) = 0 \]
This gives us one solution \( b = 0 \). To find the other solutions, we need to solve the quadratic equation:
\[ -3b^2 + 5b - 1 = 0 \]
Using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
Here, \( A = -3 \), \( B = 5 \), and \( C = -1 \):
Calculating the discriminant:
\[ B^2 - 4AC = 5^2 - 4(-3)(-1) = 25 - 12 = 13 \]
Finding \( b \):
\[ b = \frac{-5 \pm \sqrt{13}}{-6} = \frac{5 \mp \sqrt{13}}{6} \]
Thus, the solutions for \( b \) are:
\[ b = 0, \quad b = \frac{5 + \sqrt{13}}{6}, \quad b = \frac{5 - \sqrt{13}}{6} \]
These represent all values of \( b \) such that the average value of \( f(x) \) over the interval \([0, b]\) is equal to \( 4 \).
The final answer is:
\[ 0, \frac{5 + \sqrt{13}}{6}, \frac{5 - \sqrt{13}}{6} \]
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