Key Concepts and Formulas

• Lagrange's Mean Value Theorem (LMVT): If a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
• Differentiability implies Continuity: If a function is differentiable at a point, it is also continuous at that point.
• Properties of Inequalities: Multiplying or dividing both sides of an inequality by a positive number preserves the inequality.

Step-by-Step Solution

1. Analyze the Given Information We are given:

• $f(x)$ is differentiable for all $x$. This implies $f(x)$ is also continuous for all $x$.
• $f(1) = -2$
• $f'(x) \ge 2$ for $x \in [1, 6]$ Our goal is to find a lower bound for $f(6)$.

2. Verify Conditions for Applying Lagrange's Mean Value Theorem We want to apply LMVT on the interval $[1, 6]$. Let $a = 1$ and $b = 6$.

• Continuity: Since $f(x)$ is differentiable for all $x$, it is also continuous for all $x$. Thus, $f(x)$ is continuous on $[1, 6]$.
• Differentiability: We are given that $f(x)$ is differentiable for all $x$, so it is differentiable on $(1, 6)$. Both conditions are satisfied, so LMVT can be applied.

3. Apply Lagrange's Mean Value Theorem By LMVT, there exists a $c \in (1, 6)$ such that $f'(c) = \frac{f(6) - f(1)}{6 - 1}$ Simplifying, we have $f'(c) = \frac{f(6) - f(1)}{5} \quad (*)$

4. Utilize the Given Derivative Inequality We are given that $f'(x) \ge 2$ for all $x \in [1, 6]$. Since $c \in (1, 6)$, we have $f'(c) \ge 2$

5. Combine the Results to Form an Inequality Substitute $f'(c) \ge 2$ into equation $(*)$: $\frac{f(6) - f(1)}{5} \ge 2$ Multiply both sides by 5: $f(6) - f(1) \ge 10$ Substitute $f(1) = -2$: $f(6) - (-2) \ge 10$ $f(6) + 2 \ge 10$ Subtract 2 from both sides: $f(6) \ge 8$

6. Conclusion and Option Matching We have found that $f(6) \ge 8$. Comparing this to the given options: (A) $f(6) \ge 8$ (B) $f(6) < 8$ (C) $f(6) < 5$ (D) $f(6) = 5$ Our result matches option (A).

Common Mistakes & Tips

• Forgetting to check LMVT conditions: Always verify that the function is continuous on the closed interval and differentiable on the open interval before applying LMVT.
• Incorrectly manipulating inequalities: Remember to flip the inequality sign when multiplying or dividing by a negative number.
• Misunderstanding LMVT: LMVT guarantees the existence of a point $c$, but it doesn't tell you its exact value. You need to use the given information to proceed.

Summary

This problem demonstrates a direct application of the Lagrange Mean Value Theorem. We used the given information about the derivative's lower bound and the function's value at a specific point, along with LMVT, to determine a lower bound for the function's value at another point. The final inequality $f(6) \ge 8$ was derived by applying LMVT and using the given inequality.

The final answer is \boxed{f(6) \ge 8}, which corresponds to option (A).