A continuous function f, defined for all x, has the following properties:
1. f is increasing
2. f is concave down
3. f(13)=3
4. f'(13)=1/4
Sketch a possible graph for f, and use it to answer the following questions about f.
A. For each of the following intervals, what is the minimum and maximum number of zeros f could have in the interval? (Note that if there must be exactly N zeros in an interval, the minimum and maximum are both N.)
−INF <x<= 0
0 <x<=1
1<x<13
13<=x<INF
I need the maximum and minimum... I just have the two first ones... 0 for both 1 and 2 maximum and minimum... then I really don't know what to do..
B. Are any of the following possible values for f'(1)? (Enter your answer as a comma-separated list, or enter 'none' if none of them are possible.) −3, −2, −1, −51, 0, 51, 1, 2, 3.
possible values: f'(1)=_________
C. What happens to f as x−>- INF?
lim x−> INF f(x)= ________
(Enter the value, 'infinity' or '-infinity' for or −, or 'none' if there is no limit.)
I realy don't know how to do these problems.. please help
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To approach these questions, we will use the given properties of the function and the concept of derivative. Let's break down each question and explain step by step how to find the answers.
A. To find the minimum and maximum number of zeros that f could have in each interval, we need to analyze the behavior of the function based on its properties.
1. For the interval −INF < x <= 0:
Since f is increasing, it means the function is constantly rising. Therefore, there can be a minimum of zero and a maximum of one zero in this interval.
2. For the interval 0 < x <= 1:
Since f is increasing, and from the previous result, we know it can have at most one zero at x = 0. Now, as x increases from 0 to 1, f will continue to rise without any change in concavity (due to the concave-down property). This implies that the function will not cross the x-axis again in this interval. Hence, the minimum and maximum number of zeros in this interval are both zero.
3. For the interval 1 < x < 13:
Since f is increasing and concave down, it will have at least one zero in this interval due to the nature of the concavity. However, we cannot determine the maximum number of zeros based on the given properties alone.
4. For the interval 13 <= x < INF:
As f(13) = 3 and f'(13) = 1/4 > 0, we know that at x = 13, the function is still increasing. Hence, considering the concave-down property, the function will not cross the x-axis again in this interval. So the minimum and maximum number of zeros in this interval are both zero.
B. To find the possible values for f'(1), we can make use of the given information that f'(13) = 1/4. Since the function is increasing and concave down, it means the slope of the function is positive and decreasing. Therefore, f'(x) is a decreasing function.
From this information, we can conclude that f'(1) must be less than f'(13), which is 1/4. Among the given options, the possible values for f'(1) are −3, −2, −1, 0, and 1.
C. To determine what happens to f as x approaches negative infinity (x→−INF), we need to consider the behavior of the function based on its properties.
As f is increasing and concave down, it means that the function will continue to rise at a decreasing rate indefinitely as x approaches negative infinity. Therefore, the limit of f as x→−INF is −∞ (negative infinity).
In summary:
A. For each interval, the minimum and maximum number of zeros are:
−INF < x <= 0: min = 0, max = 1
0 < x <= 1: min = 0, max = 0
1 < x < 13: min = 1, max = undetermined
13 <= x < INF: min = 0, max = 0
B. Possible values for f'(1): −3, −2, −1, 0, 1
C. lim x→−INF f(x) = -∞
To address your concerns, let's go step by step:
A. To determine the minimum and maximum number of zeros for each interval, we can analyze the graph. Based on the given properties of the function, we know that:
- When the function is increasing and concave down, it will be positive and increasing but eventually tapering off.
- Since f(13) = 3, the graph must pass through the point (13, 3).
- Since f'(13) = 1/4, the graph must have a positive slope at x = 13.
Considering these properties, we can make a sketch of a possible graph:
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Now, let's determine the minimum and maximum number of zeros for each interval:
1. For −INF < x <= 0: The graph is positive and increasing, so there are no zeros. The minimum and maximum number of zeros is 0.
2. For 0 < x <= 1: The graph is positive and increasing, so there are no zeros. The minimum and maximum number of zeros is 0.
3. For 1 < x < 13: The graph is positive, increasing, but eventually tapering off. It crosses the x-axis once. The minimum and maximum number of zeros is 1.
4. For 13 <= x < INF: The graph is positive, increasing, and continues without any turning points. It never crosses the x-axis. The minimum and maximum number of zeros is 0.
Thus, we have:
- For −INF < x <= 0, there are 0 zeros.
- For 0 < x <= 1, there are 0 zeros.
- For 1 < x < 13, there is 1 zero.
- For 13 <= x < INF, there are 0 zeros.
B. To determine the possible values for f'(1), let's consider the given information. We know that f is increasing, so its derivative must be positive. Therefore, f'(1) cannot be negative.
Therefore, the possible values for f'(1) are:
None of the following values are possible: −3, −2, −1, −51, 0, 51.
C. To determine what happens to f as x approaches -INF, we can analyze the given properties. As f is an increasing and concave down function, it does not have any asymptotes. Therefore, we can say that:
lim x -> -INF f(x) = -INF
This means that as x approaches negative infinity, f(x) approaches negative infinity.
I hope this clarifies the problems for you. Let me know if you have any further questions!