Let's break down the question in simpler terms. It’s asking if we can find a smooth path, called a function $f(x)$, that has three specific rules:

1. Starting Point: The function starts at a height of 10 when $x = 0$. So, $f(0) = 10$.
2. Ending Point: The function needs to end at a height of 2 when $x = 2$. So, $f(2) = 2$.
3. Slope Rule: As we move along this path from $x = 0$ to $x = 2$, the slope (or steepness) can’t be too steep. Specifically, the slope $f'(x)$, which tells us how fast the function goes up or down, can never be more than 1.

Now, let’s imagine drawing this path on a graph. Starting at 10 and going down to 2 means it has to drop by 8 units over a distance of 2 units (from $x = 0$ to $x = 2$).

If the slope is limited to 1, that means for every step we take to the right (increasing $x$), we can only drop 1 unit. But we need to drop a total of 8 units, and we only have 2 steps (from $x = 0$ to $x = 2$) to do it.

So, if we try to drop 8 units in 2 steps, we’d need a slope of 4 (since 8 divided by 2 is 4). But the rule says the slope can’t be more than 1, which makes it impossible to reach 2 from 10 while keeping the slope under control.

Conclusion: No, it’s not possible to have such a path with these rules. The slope restriction makes it impossible to drop from 10 to 2 in just 2 units of distance.

Would you like more details on why this slope limitation creates a problem?

Here are five questions to expand on this:

1. What is the concept of a derivative, and why does it relate to the slope of a function?
2. How would relaxing the slope limit change the answer to this problem?
3. Can we solve this problem by increasing the interval distance?
4. What happens if we change the values of $f(0)$ or $f(2)$?
5. How would the answer change if we allowed the function to be discontinuous?

Tip: Think of the slope as the "steepness" of a hill. If the hill is too steep for you to safely descend at a controlled rate, you’ll go down faster than allowed!