Can someone help me please!!!
1 answer:
Answer:
Step-by-step explanation:
Given: A town's population has been growing linearly. In 2003, the population was 35000, and the population has been growing by 1200 (constant rate of change) people each year.
The equation of a line is given by :-
, where m is the slope ( constant rate of change) and c is the initial value of the function.
From the given statement , we have the equation for the population x years after 2003 :-
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Answer:
a)
x = -1/√3 = -0.58 is a local maximum
x = 1/√3 = 0.58 is a local minimum
b)
(-1/√3 , 1/√3 ) DECREASING
(-∞, -1/√3) U (1/√3, ∞) INCREASING
Step-by-step explanation:
To answer all of the questions we must obtain the derivative of the fucntion:
If
U(x) = 4(x^3 - x)
then
U'(x) = 4(3x^2 - 1)
U''(x) = 4(3*2 x) = 24 x
U'''(x) = 24
The local maxima and minima of the function U(x) can be found when U'(x) = 0
this occurs when :
3x^2 = 1
that is:
x = ±1/√3
We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as <u><em>Second derivative test</em></u>), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.
Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.
a)
x = -1/√3 = -0.58 is a local maximum
x = 1/√3 = 0.58 is a local minimum
b)
the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }
The first derivative will be positive when:
3x^2 > 1
|x| > 1/√3 --> x > 1/√3 and x < -1/√3
The first derivatice will be negative when:
3x^2 < 1
|x| < 1/√3 --> x < 1/√3 and x > -1/√3
Therefore the intervals are:
(-1/√3 , 1/√3 ) DECREASING
(-∞, -1/√3) U (1/√3, ∞) INCREASING
<u><em>** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum</em></u>
