To solve this problem, we need to estimate the maximum possible value of $f(6)$ given the following information:

1. $f(2) = 6$.
2. The derivative $f'(x) \leq 5$ for all values of $x$.

The key concept here is to use the Mean Value Theorem (MVT), which states that for a continuous function $f$ on the interval $[a, b]$ that is differentiable on $(a, b)$, there exists a point $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}.$

Step-by-Step Solution

We want to determine the maximum possible value of $f(6)$, starting from $f(2) = 6$.

1. Apply MVT over the interval $[2, 6]$: By the Mean Value Theorem, there exists a point $c$ in $(2, 6)$ such that $f'(c) = \frac{f(6) - f(2)}{6 - 2}.$

2. Substitute known values: Since $f(2) = 6$, we get $f'(c) = \frac{f(6) - 6}{4}.$

3. Use the derivative constraint $f'(x) \leq 5$: Since $f'(x) \leq 5$ for all $x$, we can assume $f'(c) \leq 5$. Thus, $\frac{f(6) - 6}{4} \leq 5.$

4. Solve for $f(6)$: Multiply both sides by 4 to obtain $f(6) - 6 \leq 20.$ Adding 6 to both sides, we find $f(6) \leq 26.$

Conclusion

The largest possible value of $f(6)$ is 26.

Would you like more details on any step, or do you have further questions?

Related Questions:

1. How would the answer change if $f'(x) \leq 3$ instead of $f'(x) \leq 5$?
2. Can the Mean Value Theorem be used if $f(x)$ is not differentiable on $(2,6)$?
3. What if we knew $f(2) = 6$ and $f(4) = 16$? Could we find $f'(x)$ for some $x$?
4. How does the Mean Value Theorem help in real-world applications?
5. How does the assumption of $f'(x) \leq 5$ affect the function's behavior on the interval?

Tip:

The Mean Value Theorem is a powerful tool to understand how much a function can change over an interval when there are constraints on its derivative.