Find x* on [0, 2] such that f(x*) equals the average value of f(x) = 5x + 9 over [0, 2].
Tha average value of f is the integral of the function (over that [0,2] interval) divided by the interval width, 2.
I assume you know how to integrate the function. The value of the integral is
5/2*4 + 18 = 28 . The average value is therefre 28/2 = 14
Now find what value of x that makes
5x + 9 = 14. It is 1
That should not be a surprise. For a linear function, the average is the value in the middle of the interval.
To find the value of x* such that f(x*) equals the average value of f(x) over [0, 2], we first need to find the average value of f(x).
The average value of a function f(x) over an interval [a, b] is given by the formula:
Average value = (1 / (b - a)) * ∫[a, b](f(x) dx)
In this case, we need to find the average value of f(x) = 5x + 9 over the interval [0, 2].
Now, let's calculate the integral:
∫[0, 2] (5x + 9) dx = [5/2 * x^2 + 9x] from 0 to 2
= (5/2 * 2^2 + 9 * 2) - (5/2 * 0^2 + 9 * 0)
= (5/2 * 4 + 9 * 2) - (0 + 0)
= (20/2 + 18) - 0
= 30
Next, let's calculate the length of the interval [0, 2]:
Length of interval = 2 - 0 = 2
Now, we can find the average value:
Average value = (1 / (2 - 0)) * 30
= 1/2 * 30
= 15
We want to find the value of x* such that f(x*) equals the average value of 15.
So, now we have the equation: f(x*) = 15
Substituting f(x) = 5x + 9 into the equation, we get:
5x* + 9 = 15
Solving for x*, we have:
5x* = 15 - 9
x* = (15 - 9) / 5
x* = 6 / 5
x* = 1.2
Therefore, the value of x* such that f(x*) equals the average value of f(x) = 5x + 9 over [0, 2] is x* = 1.2.