This problem involves the Mean Value Theorem (MVT) and understanding how the derivative affects the behavior of the function. Let’s break it down:

Mean Value Theorem (MVT):

The MVT states that if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}.$

Given Information:

• $f(1) = 10$,
• $f'(x) \geq 2$ for $1 \leq x \leq 4$.

We need to find the smallest possible value of $f(4)$.

Step-by-Step Solution:

1. MVT Application:

   • For $f$ on the interval $[1, 4]$, MVT guarantees there exists some $c \in (1, 4)$ such that: $f'(c) = \frac{f(4) - f(1)}{4 - 1}.$

   • Substituting $f(1) = 10$: $f'(c) = \frac{f(4) - 10}{3}.$

2. Bounding $f'(c)$:

   • Since $f'(x) \geq 2$ for all $x \in [1, 4]$, it follows that: $\frac{f(4) - 10}{3} \geq 2.$

3. Solve for $f(4)$:

   • Multiply through by 3: $f(4) - 10 \geq 6.$

   • Add 10 to both sides: $f(4) \geq 16.$

Final Answer:

The smallest possible value of $f(4)$ is: $\boxed{16}.$

Would you like further clarification on MVT or how the inequality was derived? Let me know!

Related Questions:

1. What is the Mean Value Theorem and how does it apply to differentiable functions?
2. If $f'(x) > 2$, how would the minimum value of $f(4)$ change?
3. How do we interpret $f'(x) \geq 2$ geometrically for the function $f(x)$?
4. What additional constraints on $f'(x)$ could impact $f(4)$'s minimum value?
5. Could we use a different theorem (e.g., Rolle's Theorem) in this scenario, and why or why not?

Tip:

Always visualize the problem with a graph when working with derivatives and inequalities—it helps to understand the behavior of the function.