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

A. To determine the maximum and minimum number of zeros in each interval, we can look at the behavior of the graph of f. Since f is increasing and concave down, it means that the graph of f will be a curve that rises continuously and curves downward.

1. For the interval −INF < x ≤ 0: Since f is increasing, it means that it will not have any zeros in this interval. So, the minimum and maximum number of zeros are both 0.

2. For the interval 0 < x ≤ 1: Again, since f is increasing, it will not have any zeros in this interval. Minimum and maximum number of zeros are both 0.

3. For the interval 1 < x < 13: In this interval, the graph curves downward but is still increasing. Therefore, there could be minimum and maximum of 1 zero in this interval.

4. For the interval 13 ≤ x < INF: As we know that f(13) = 3, and f'(13) = 1/4, it tells us that the graph of f is increasing and becomes flatter near x = 13. So, there will still be a single zero in this interval. Minimum and maximum number of zeros are both 1.

B. To determine the possible values of f'(1), we need to know more about the behavior of f and its derivative. Given the information we have, it's difficult to determine the exact value of f'(1). So, it's possible to have any of the given values for f'(1): −3, −2, −1, −51, 0, 51, 1, 2, 3.

C. As x → -∞, it means x is approaching negative infinity, and we need to determine the behavior of f in that direction. Since f is increasing, it means that as x becomes more and more negative, the values of f(x) will also become more and more negative. So, we can say that as x → -∞, lim x→-∞ f(x) = -∞ (negative infinity).

To answer these questions, let's consider the properties given and use them to sketch a possible graph for f.

Based on the given properties:
1. f is increasing: This means that as x increases, f(x) also increases.
2. f is concave down: This means that the graph of f will be curving downwards.
3. f(13)=3: This gives us a specific point on the graph, (13, 3).
4. f'(13)=1/4: This tells us about the slope of the tangent line at x=13.

Using these properties, we can sketch a possible graph as follows:

First, draw a straight line with a positive slope passing through (13, 3). This represents the increasing property of f.

Next, make the graph curve downwards (concave down). The exact shape of the curve isn't specified, but we know it should be smooth and continuously decreasing.

Now let's answer the questions:

A. For each of the following intervals, what is the minimum and maximum number of zeros f could have in the interval?

-∞ < x ≤ 0: Since f is increasing, the minimum and maximum number of zeros f could have in this interval is 0.
0 < x ≤ 1: The same reasoning applies, so the minimum and maximum number of zeros f could have in this interval is also 0.

Now let's look at the other intervals:

1 < x < 13: Since f is increasing, there must be at least one zero in this interval. However, because f is concave down, it can have at most two zeros in this interval. So the minimum number of zeros is 1 and the maximum number of zeros is 2.

13 ≤ x < ∞: Since f is increasing, there must be at least one zero in this interval. But since we don't know how the graph behaves for large x values, we can't determine the maximum number of zeros in this interval. So the minimum number of zeros is 1, but the maximum number is unknown.

B. Are any of the following possible values for f'(1)? −3, −2, −1, −51, 0, 51, 1, 2, 3.

Since f is increasing, the slope of the tangent line at any point should be positive or zero. Therefore, the only possible value for f'(1) is 0.

C. What happens to f as x → -∞?

As x approaches negative infinity, the graph of f will continue to increase indefinitely because f is an increasing function. Therefore, the limit as x approaches -∞ is infinity.

To solve these problems, we need to use the information provided about the properties of the function f.

A. For each interval, we will analyze the number of zeros that f could have:

1. −∞ < x ≤ 0: Since f is increasing and concave down, and f(13) = 3, the graph of f must start from below the x-axis (negative y-values) before x = 0 and then cross the x-axis at some point after x = 0. Therefore, the minimum and maximum number of zeros in this interval is 1.

2. 0 < x ≤ 1: Similarly, the graph of f must cross the x-axis at some point between x = 0 and x = 1. Therefore, the minimum and maximum number of zeros in this interval is also 1.

3. 1 < x < 13: Since f is increasing and concave down, the graph of f will not cross the x-axis in this interval.

4. 13 ≤ x < ∞: Since f(13) = 3 and f'(13) = 1/4, we know that the graph of f will approach x = 13 from below and continue to increase with a positive slope. Therefore, the graph of f will not cross the x-axis in this interval either.

In summary, the answers for A are:
1. −∞ < x ≤ 0: Minimum = Maximum = 1 zero
2. 0 < x ≤ 1: Minimum = Maximum = 1 zero
3. 1 < x < 13: Minimum = Maximum = 0 zeros
4. 13 ≤ x < ∞: Minimum = Maximum = 0 zeros

B. To find the possible values for f'(1), we can use the fact that f is increasing and concave down. Since f(13) = 3 and f'(13) = 1/4, as x increases from 13, f'(x) will be positive and decreasing. This means that f'(1) must be greater than 1/4. Therefore, the only possible value for f'(1) is 51.

C. To find the limit of f as x approaches −∞, we need to consider the behavior of the graph of f. Since f is increasing and concave down, and f(13) = 3, the graph of f will approach +∞ as x approaches −∞. Therefore, the limit of f as x → −∞ is +∞.

In summary:
A. 1. −∞ < x ≤ 0: Minimum = Maximum = 1 zero
2. 0 < x ≤ 1: Minimum = Maximum = 1 zero
3. 1 < x < 13: Minimum = Maximum = 0 zeros
4. 13 ≤ x < ∞: Minimum = Maximum = 0 zeros

B. Possible value for f'(1) is 51.

C. lim x → −∞ f(x) = +∞.

For a function to be increasing on ℝ and concave downwards, it must:

1. have no maximum/minimum, because it is a one-to-one function.
2. Crosses the x-axis (or any other y-value) at most once, since it is an increasing function.
3. Since it is concave down, d²y/dx² must be negative.

Study the above statements. If you do not know why they are true, refer to your notes, your textbook, or post.

I have taken an example function that satisfies these two properties is f(x)=3-e^(-x). (There may be many other functions that have these properties.) See the following link for its graph:
http://img268.imageshack.us/i/1298419029.png/

A. Since we know that f(13)>0, then the zero of the function, if any, must happen when x<13 for an increasing function. Beyond 13, x continues to increase, and therefore cannot have zeroes.

B. If f'(13)=1/4, and we know that f"(x) is negative throughout the domain of f(x), what can you say about f'(x) for x>13? Would it not be true that f'(x)<1/4 for x>13?
So what happens to to f'(x) <13?

C. The limit for the example function is 3, (but f(x) does not satisfy the required conditions of f(13)=3 and f'(13)=1/4.) Some parameters are required. It was just for illustration purposes.