Which is an irrational number?
1 answer:
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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).
Answer:
A. Increase by 2
Step-by-step explanation:
Given that a fitted multiple regression equation is
This is a multiple regression line with dependent variable y and independent variables x1, x2, x3 and x4
The coefficients of independent variables represent the slope.
In other words the coefficients represent the rate of change of y when xi is changed by 1 unit.
Given that x3 and x4 remain unchanged and x1 increases by 2 and x2 by 2 units
Since slope of x1 is 5, we find for one unit change in x1 we can have 5 units change in y
i.e. for 2 units change in x1, we expect 10 units change in Y
Similarly for 2 units change in x2, we expect -2(4) units change in Y
Put together we have
change in y
Since positive 2, there is an increase by 2
A. Increase by 2