State two intervals in which the function y=7sin(1/5x)+2 has an average rate of change that is:
a) zero
b)a negative value
c)a positive value
Please explain how to get the intervals! Thanks a lot in advance!
a) pick two values of x where y is the same. The average rate of change for that interval will be zero.
For example, if x = pi, then y = 7sin(pi/5)+2
Now sin(4pi/5) = sin(pi/5), so if x = 4pi, then
y = 7sin(4pi/5) + 2
so, on the interval [pi,4pi] the average rate of change is zero.
Taking the easy way out, just tweak the ends of the interval to make the slop of the line negative or positive.
Since sin(pi/5+.01) > sin(pi/5)
the slope on [pi+.05,4pi] will be negative
similarly, on [pi-.05,4pi], the slope is positive
a) To find the intervals where the average rate of change of a function is zero, we need to identify the points on the graph where the function is increasing and decreasing.
In this case, the given function is y = 7sin(1/5x) + 2.
Let's look at the graph of this function:
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The graph of the function resembles a sinusoidal wave. It oscillates above and below the x-axis.
To find the intervals where the average rate of change is zero, we need to identify the points where the graph crosses the x-axis.
The function y = 7sin(1/5x) + 2 crosses the x-axis at two points due to the "+2" term. The y-intercept is at (0, 2).
To find the intervals where the graph crosses the x-axis, let's analyze the behavior of the sine function. The sine function crosses the x-axis at regular intervals.
Since 7sin(1/5x) oscillates between 7 and -7, it will cross the x-axis when 7sin(1/5x) = -7.
Solving for sin(1/5x) = -1, we find that 1/5x = -π/2 + 2πk, where k is an integer.
Simplifying further, x = (-2π/5) + 10πk, where k is an integer.
Thus, the function crosses the x-axis at x = (-2π/5) + 10πk, where k is an integer.
Since we're looking for intervals, we want to find two consecutive values of k that correspond to different intervals.
Let's choose k = 0 and k = 1.
For k = 0, x = (-2π/5) + 10π(0) = -2π/5.
For k = 1, x = (-2π/5) + 10π(1) = 8π/5.
Therefore, the intervals where the average rate of change of the function y = 7sin(1/5x) + 2 is zero are (-2π/5, 8π/5).
b) To find the intervals where the average rate of change of a function is negative, we need to identify the points on the graph where the function is decreasing.
In this case, the given function is y = 7sin(1/5x) + 2.
From the graph in the previous explanation, we can see that the function is decreasing when the graph is below the x-axis.
Since the function oscillates above and below the x-axis, we need to identify the intervals where the function is below the x-axis.
The function y = 7sin(1/5x) + 2 is below the x-axis when 7sin(1/5x) + 2 < 0.
If we subtract 2 from both sides of the inequality, we get 7sin(1/5x) < -2.
Now, let's analyze the behavior of the sine function. The sine function ranges from -1 to 1.
To find the intervals where 7sin(1/5x) < -2, we need to find intervals where sin(1/5x) < -2/7.
However, there are no values of x that satisfy sin(1/5x) < -2/7 since the range of the sine function is -1 to 1.
Therefore, there are no intervals where the average rate of change of the function y = 7sin(1/5x) + 2 is negative.
c) Since we already found the intervals for part (a), we can use those intervals for part (c) as well. The intervals for part (c) are (-2π/5, 8π/5).
To find the intervals where the average rate of change of the function is zero, negative, or positive, we need to analyze the behavior of the function and identify specific points.
Let's analyze the function y = 7sin(1/5x) + 2.
a) Average Rate of Change = 0:
The average rate of change is zero when the function does not increase or decrease over an interval. This means that the function should have points where it intersects the x-axis or reaches local maximum or minimum points.
To find the intervals where the average rate of change is zero, we need to find x-values where y = 0.
For y = 7sin(1/5x) + 2 to be equal to 0, we have:
7sin(1/5x) + 2 = 0
Subtracting 2 from both sides:
7sin(1/5x) = -2
Dividing by 7:
sin(1/5x) = -2/7
Now, to solve for x, we need to take the inverse sine (sin^-1) of both sides. However, we need to be cautious as there may be multiple solutions due to the periodic nature of the sine function.
Therefore, the intervals where the average rate of change is zero are determined by the values of x that solve the equation sin(1/5x) = -2/7.
b) Average Rate of Change < 0:
The average rate of change is negative when the function is decreasing over an interval. To find the intervals where the average rate of change is negative, we need to examine the behavior of the function and identify where it is decreasing.
Analyzing the function y = 7sin(1/5x) + 2, we can conclude that the function is decreasing from the highest point to the lowest point of each sinusoidal wave. The highest point of each wave is when sin(1/5x) = 1, and the lowest point is when sin(1/5x) = -1.
Therefore, the intervals where the average rate of change is negative are determined by the values of x that solve the equations sin(1/5x) = 1 and sin(1/5x) = -1.
c) Average Rate of Change > 0:
Similarly, the average rate of change is positive when the function is increasing over an interval. To find the intervals where the average rate of change is positive, we need to examine the behavior of the function and identify where it is increasing.
Analyzing the function y = 7sin(1/5x) + 2, we can conclude that the function is increasing from the lowest point to the highest point of each sinusoidal wave.
Therefore, the intervals where the average rate of change is positive are determined by the values of x that do not solve the equations sin(1/5x) = 1 and sin(1/5x) = -1.
Please note that the exact numerical values of the intervals will depend on the specific range of x-values you are considering in your analysis.
To find intervals in which a function has a certain average rate of change, we need to examine the behavior of the function.
a) For an average rate of change of zero, the function needs to be constant within an interval. This means there should be no increase or decrease in the function values. In the case of the function y = 7sin(1/5x) + 2, the first step is to set the derivative of the function equal to zero:
dy/dx = 7/5 cos(1/5x) = 0
Solving this equation for cos(1/5x) = 0, we find that cos(1/5x) is zero at every odd multiple of π/2. Therefore, we can conclude that the function y = 7sin(1/5x) + 2 has an average rate of change of zero within the intervals [2nπ, (2n + 1)π] and [(2n + 1)π, (2n + 2)π], where n is an integer.
b) For a negative average rate of change, the function needs to exhibit a decreasing behavior within an interval. In other words, the function values should be decreasing as we move from left to right. In the case of y = 7sin(1/5x) + 2, analyze the sign of the derivative:
dy/dx = 7/5 cos(1/5x)
Since cos(1/5x) is always between -1 and 1, the derivative can be negative if cos(1/5x) is negative. This occurs in the interval [(4n + 1)π, (4n + 2)π], where n is an integer.
c) For a positive average rate of change, the function should exhibit an increasing behavior within an interval. In other words, the function values should be increasing as we move from left to right. Analyzing the sign of the derivative:
dy/dx = 7/5 cos(1/5x)
Since cos(1/5x) is always between -1 and 1, the derivative can be positive if cos(1/5x) is positive. This happens in the interval [(4n + 3)π, (4n + 4)π], where n is an integer.
In summary:
a) The function has an average rate of change of zero within the intervals [2nπ, (2n + 1)π] and [(2n + 1)π, (2n + 2)π].
b) The function has an average rate of change that is negative within the intervals [(4n + 1)π, (4n + 2)π].
c) The function has an average rate of change that is positive within the intervals [(4n + 3)π, (4n + 4)π].