Answer:
the exact answer is x = -∛2.
Explanation:
The average value of a function over an interval [a, b] is given by:
Avg = (1 / (b - a)) * ∫[a to b] f(x) dx
In this case, the interval is [−2, 0], so a = -2 and b = 0.
Avg = (1 / (0 - (-2))) * ∫[-2 to 0] (9x^3) dx
= (1 / 2) * ∫[-2 to 0] (9x^3) dx
= (1 / 2) * [9/4 * x^4] | [-2 to 0]
= (1 / 2) * (9/4 * (0^4 - (-2)^4))
= (1 / 2) * (9/4 * (0 - 16))
= (1 / 2) * (9/4 * (-16))
= (1 / 2) * (-36)
= -18
The average value of f(x) = 9x^3 over the interval [-2, 0] is -18.
Now, we need to find the points in the interval where the function takes on this average value of -18.
Setting f(x) = -18, we have:
9x^3 = -18
Dividing both sides by 9:
x^3 = -2
Taking the cube root of both sides:
x = -∛2
Therefore, the point cc in the interval [-2, 0] where f(x) = 9x^3 takes on its average value of -18 is x = -∛2.
i hope i helped!