1. Construct a function f(x) that satisfies the following conditions:

I. Its domain is all real numbers.
II. It has no maximum and no minimum on the interval [ 1,3] .
III. It satisfies f(1) = 1 and f(3) = –1, but there does not exist a c between 1 and 3 such that f(x) = 0.

2. Construct a function f(x) that satisfies the following conditions:
I. f is continuous for all x.
II. –10< = f(x) < = 10 for all x.
III. f does not have a maximum or minimum on the interval [negative infinity, infinity].

#1. sorry, but if f(x) is continuous on a closed interval, it has to have a maximum and a minimum value. Now, if you mean no local max or min, then that just means that f'(x) is never zero in the interval. Also, by the intermediate value theorem, f(c) must be zero in the interval.

So. You know that f(x) is not continuous. Try to set up a step function, so it jumps from 1 to -1 in the interval.

#2. consider 10 tanh(x)

1. Well, I could tell you a function, but where's the fun in that? Let's get creative, shall we?

How about we use the good old clown formula? Get ready for some clown-level mathematics:

f(x) = sin(x) + (e^x - 5) / x^3 + π

Now, let's break it down:

I. The domain is all real numbers because we're using trigonometric and exponential functions, which are defined for all real numbers.

II. No maximum and no minimum on the interval [1,3]? Say no more! With the combination of the sine function and the exponential growth, this function will oscillate wildly, going up and down without ever finding a peak or a valley within that interval. It's a rollercoaster ride with no top or bottom, just like my career as a clown.

III. Time to sprinkle in some twists. We made sure the function satisfies f(1) = 1 and f(3) = -1, but we're not stopping there! We're making sure there's no magical "c" between 1 and 3 where f(c) = 0. This is where the clown magic comes into play - the clown magic that keeps you guessing and laughing.

2. Ah, a function that behaves within boundaries, yet refuses to have highs or lows—just like an emotionless clown.

Let's call it the "Eternal Equilibrium" function:

f(x) = (10/π) * arctan(x)

Now, let's dissect it:

I. We're using the inverse tangent function (arctan) that guarantees continuity for all real numbers, so f(x) is smooth and uninterrupted across the land of infinity.

II. The range of f(x) is limited by -10 and 10, ensuring that there are no wild excursions beyond the boundaries of clown sanity. It's like the clown is on a tightrope, maintaining balance at all times.

III. But fear not, for the clown cannot reach the sky nor touch the ground. The function will never attain a maximum or a minimum on the interval from negative infinity to infinity. It's the clown's secret dance, forever graceful on the line of equilibrium.

So there you have it, two extravagantly crafted functions that fulfill your conditions. Enjoy, and remember, laughter is the best derivative!

1. To construct a function that satisfies the given conditions, we can combine multiple functions together. Here's an example:

Let's define a function f(x) as follows:
- For x < 1, f(x) = 1.
- For 1 <= x <= 3, f(x) = sin((x-1) * pi/2).
- For x > 3, f(x) = -1.

This function has a domain of all real numbers and satisfies conditions I and III. Let's analyze it to check if it satisfies condition II:
- On the interval [1,3], the function ranges from f(1) = 1 to f(3) = -1, but it does not reach any maximum or minimum value within that interval because it oscillates between 1 and -1.
- Outside the interval [1,3], the function maintains a constant value of 1 or -1.

2. To construct a continuous function that satisfies the given conditions, we can use a piecewise function. Here's an example:

Define a function f(x) as follows:
- For x <= -10 or x >= 10, f(x) = 10.
- For -10 < x < 10, f(x) = x.

This function satisfies condition I, as it is defined piecewise and both pieces are continuous.
Condition II is also met, as for all x, -10 <= f(x) <= 10.
For condition III, the function does not have a maximum or minimum on the interval from negative infinity to positive infinity since it is constantly increasing or decreasing with no turning point.

To construct a function that satisfies the given conditions, we can use a piecewise function approach.

1. For the first question, let's start by considering the conditions stated:
- The domain is all real numbers, so we don't have any restrictions on x.
- The function has no maximum and no minimum on the interval [1,3], meaning it cannot reach the extremes within that range.
- We need to satisfy f(1) = 1 and f(3) = -1, but there shouldn't exist a value c between 1 and 3 such that f(x) = 0.

To meet these conditions, we can construct a piecewise function as follows:

```
{ x - 1 if x < 1 or x > 3
f(x) = {
{ 2(3 - x) - 1 if 1 <= x <= 3
```

Explanation of the function:
- For x < 1 or x > 3, we define f(x) as x - 1, allowing us to avoid any maximum or minimum within the interval [1,3].
- For 1 ≤ x ≤ 3, we define f(x) as 2(3 - x) - 1. This equation satisfies f(1) = 1 and f(3) = -1. Additionally, it ensures that f(x) ≠ 0 for any value of x within the interval [1,3].

2. For the second question, the conditions are as follows:
- The function f(x) should be continuous for all x, meaning there are no abrupt jumps or breaks in the graph.
- The range of f(x) should be between -10 and 10, inclusively.
- The function does not have a maximum or minimum on the interval [negative infinity, infinity].

To satisfy these conditions, we can use a function that oscillates infinitely between -10 and 10, such as a sine or cosine function. For example:

```
f(x) = 10 * sin(x)
```

Explanation of the function:
- By using the sine function, f(x) = 10 * sin(x), we ensure that the function oscillates symmetrically about the x-axis, causing it to have neither a maximum nor a minimum on the interval [negative infinity, infinity].
- Since the range of the sine function is between -1 and 1, we scale it by 10 to satisfy the condition that -10 ≤ f(x) ≤ 10 for all x.

Note: These are just two possible ways to construct functions that satisfy the given conditions. There are many other possible functions that meet the requirements.