The average value of the function f(x) = 3x^2 is 3 on the inetrval [-3, 3].
According to the given question.
We have a function.
F(x) = 3x^2
Since, we know that " the average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval".
Here, the given interval is [-3, 3]
Therefore, the length of the interval = 3 - (-3) = 3 + 3 = 6
Now, the average value of the given function f(x)

![= \frac{1}{6} [\frac{x^{3} }{3} ]_{3} ^{-3}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B6%7D%20%5B%5Cfrac%7Bx%5E%7B3%7D%20%7D%7B3%7D%20%5D_%7B3%7D%20%5E%7B-3%7D)


= 2(27)/18
= 27/9
= 3
Hence, the average value of the function f(x) = 3x^2 is 3 on the inetrval [-3, 3].
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Answer:
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Answer:
I assume the answer would be that f(x) = 6(x) increases over the interval x = 3 to x = 4
Step-by-step explanation:
6 x 3 is 18 and 6 x 4 is 24, so over the interval x = 3 to x = 4 f(x) would increase.
Answer:
a) 7
Step-by-step explanation:
a+b = 12
b+c = 17
a+b+c=22
12+c=22
c=10
b=7
a=5