To determine whether such a function $f$ exists, let us analyze the problem step by step:

Problem Constraints:

1. $f(0) = -1$
2. $f(2) = 6$
3. $f'(x) \leq 2$ for all $x$ (the derivative condition restricts the slope of $f(x)$).

The key is to check whether the conditions on $f'(x)$ allow the function $f$ to increase from $-1$ at $x = 0$ to $6$ at $x = 2$, given that the maximum slope of the function is 2.

Step 1: Maximum possible increase

The derivative $f'(x) \leq 2$ means that the rate of change of $f$ is at most 2. If $f$ increases at this maximum rate (constant slope of 2), then: $f(x) = f(0) + 2x.$ At $x = 2$, the value of $f(x)$ would be: $f(2) = -1 + 2(2) = 3.$

This is less than the required $f(2) = 6$. Therefore, no such function $f$ exists that satisfies all three conditions simultaneously.

Conclusion:

There does not exist a function $f$ such that $f(0) = -1$, $f(2) = 6$, and $f'(x) \leq 2$ for all $x$.

Would you like further explanation or additional examples? Here are some questions to deepen your understanding:

1. Why does the derivative constraint restrict the maximum change of a function?
2. What happens if the derivative condition $f'(x) \leq 2$ is replaced by $f'(x) \leq 3$?
3. How does the Mean Value Theorem help us analyze such problems?
4. Can we construct a piecewise function that "almost" satisfies the given conditions?
5. How would the solution change if $f'(x)$ could exceed 2 for some $x$?

Tip: Use the Mean Value Theorem as a tool to link function values to derivative constraints in problems involving slope restrictions.