Find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 9.
f(x)=7+10x-3x^2
average value*b= INT f(x) dx from 0 to b
integrate, then divide by b.
To find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 9, we can use the concept of average value and the definite integral.
The average value of a function f(x) on an interval [a, b] can be found using the formula:
Avg = 1 / (b - a) * ∫[a to b] f(x) dx
In this case, we want the average value to be equal to 9. So we have:
9 = 1 / (b - 0) * ∫[0 to b] (7 + 10x - 3x^2) dx
To solve this equation and find the value(s) of b, we need to integrate the function f(x) with respect to x and set up the equation:
9 = 1 / b * ∫[0 to b] (7 + 10x - 3x^2) dx
Integrating the function f(x) term by term, we have:
9 = 1 / b * [7x + 5x^2 - x^3/3] evaluated from 0 to b
Now, we substitute the upper limit b into the integrated function:
9 = 1 / b * [7b + 5b^2 - b^3/3] - [0]
Simplifying the equation further, we have:
9 = 7 + 5b - b^2/3
Rearranging the terms and multiplying through by 3 to eliminate the fraction, we get:
0 = b^2 - 15b + 18
Now we have a quadratic equation in terms of b. We can solve it by factoring, completing the square, or using the quadratic formula.
Factoring the quadratic equation, we get:
0 = (b - 3)(b - 6)
This gives us two possible values for b: b = 3 and b = 6.
Therefore, the numbers b such that the average value of f(x) on the interval [0, b] is equal to 9 are b = 3 and b = 6.