Question

Find all numbers b such that the average value of
f(x) = 3 + 10x − 9x2
on the interval
[0, b]
is equal to 4. (Enter your answers as a comma-separated list.)

Answer 1

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:

1. \(\int_0^b 3 \, dx = 3b\)
2. \(\int_0^b 10x \, dx = [5x^2]_0^b = 5b^2\)
3. \(\int_0^b 9x^2 \, dx = [3x^3]_0^b = 3b^3\)

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:

\[
0, \frac{5 + \sqrt{13}}{6}, \frac{5 - \sqrt{13}}{6}
\]