Find the average value of the function over the given interval and all values of x in the interval for which the function equals its average value.
f(x) = 4x3 − 3x2, [−1, 2]
see the related questions below. Just use your function instead.
Didn't work
boy, you don't give much away, do you?
The average value is
∫[a,b] f(x) dx
------------------
b-a
So, what didn't work?
∫[-1,2] 4x^3-3x^2 dx
= x^4-x^3 [-1,2]
= (16-8)-(1+1)
= 6
6/3 = 2
So, where is f(x)=2?
4x^3-3x^2=2
x ≈ 1.137
Ariel has one gallon container of juice and one quart container of juice. how many 1-cup servings of juice does Ariel have?
To find the average value of a function over a given interval, you can use the formula:
Average value = (1 / (b - a)) * ∫(a to b) f(x) dx
where a and b are the lower and upper limits of the interval, and ∫(a to b) f(x) dx is the definite integral of the function over the interval.
In this case, the function is f(x) = 4x^3 - 3x^2, and the interval is [−1, 2]. So, let's calculate the average value.
Average value = (1 / (2 - (-1))) * ∫(-1 to 2) (4x^3 - 3x^2) dx
First, let's find the indefinite integral of the function:
∫(4x^3 - 3x^2) dx = x^4 - x^3
Next, we'll evaluate the definite integral by substituting the limits:
∫(-1 to 2) (4x^3 - 3x^2) dx = [(2^4 - 2^3) - ((-1)^4 - (-1)^3)]
Simplifying further:
∫(-1 to 2) (4x^3 - 3x^2) dx = (16 - 8) - (1 + 1) = 8 - 2 = 6
Now, we can calculate the average value:
Average value = (1 / (2 - (-1))) * 6 = (1 / 3) * 6 = 2
So, the average value of the function over the interval [−1, 2] is 2.
To find the values of x in the interval for which the function equals its average value, we need to solve the equation f(x) = 2:
4x^3 - 3x^2 = 2
We can rearrange the equation to:
4x^3 - 3x^2 - 2 = 0
Unfortunately, this equation cannot be easily solved algebraically. However, we can use numerical methods like a graphing calculator or software to find the approximate values of x that satisfy the equation.