A dance school allows a maximum of 15 students per class of 112 students signing up for class how many classes does the school n
2 answers:
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Answer:
Where 0 < x < 3
The location of the local minimum, is (2, 0)
The location of the local maximum is at (0, 16)
Step-by-step explanation:
The given function is f(x) = (x + 2)⁴
The range of the minimum = 0 < x < 3
At a local minimum/maximum values, we have;
∴ (-x + 2)³ = 0
x = 2
When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2
We have, f(2) = (-2 + 2)⁴ = 0
The location of the local minimum, is (2, 0)
Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3
When x = 0, -x + 2 = 0 + 2 = 2
Similarly, we have;
-x + 2 = 1, when x = 1
-x + 2 = 0, when x = 2
-x + 2 = -1, when x = 3
Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16
The location of the local maximum is at (0, 16).
First, you should solve for 
, which equals 
. Now, solve the integral of =, to get that
. You can check this by taking the integral of what you got. Now by the Fundamental Theorem![\int\limits^2_0 {4x} \, dx=[2x^2] ^{2}_{0}=2(2)^{2}-2(0)^2=8](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B4x%7D%20%5C%2C%20dx%3D%5B2x%5E2%5D%20%5E%7B2%7D_%7B0%7D%3D2%282%29%5E%7B2%7D-2%280%29%5E2%3D8)
.
This should be the answer to your question, if I understood what you were asking correctly.
This should be the answer to your question, if I understood what you were asking correctly.