Key Concepts and Formulas

• Lagrange's Mean Value Theorem (LMVT): If a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
• The given condition $f'(x) \le 2$ provides an upper bound on the rate of change of the function.
• Using LMVT, we can relate the function values at different points to the derivative, allowing us to find the range of possible values for $f(-1)$ and $f(0)$.

Step-by-Step Solution

Step 1: Apply LMVT on the interval [-7, -1].

• Why: We want to find a bound for $f(-1)$, given $f(-7)$ and $f'(x)$. Applying LMVT on [-7, -1] will relate $f(-1)$ and $f(-7)$ using $f'(c)$ for some $c$ in (-7, -1).
• Since $f$ is continuous on [-7, 0] and differentiable on (-7, 0), it satisfies the conditions for LMVT on the interval [-7, -1]. Therefore, there exists a $c_1 \in (-7, -1)$ such that: $f'(c_1) = \frac{f(-1) - f(-7)}{-1 - (-7)} = \frac{f(-1) - f(-7)}{6}$
• We are given that $f'(x) \le 2$ for all $x \in (-7, 0)$. Thus, $f'(c_1) \le 2$. Substituting $f(-7) = -3$: $\frac{f(-1) - (-3)}{6} \le 2$ $f(-1) + 3 \le 12$ $f(-1) \le 9$

Step 2: Apply LMVT on the interval [-7, 0].

• Why: Similar to Step 1, we want to find a bound for $f(0)$, given $f(-7)$ and $f'(x)$.
• Applying LMVT on the interval [-7, 0], there exists a $c_2 \in (-7, 0)$ such that: $f'(c_2) = \frac{f(0) - f(-7)}{0 - (-7)} = \frac{f(0) - f(-7)}{7}$
• Since $f'(x) \le 2$ for all $x \in (-7, 0)$, we have $f'(c_2) \le 2$. Substituting $f(-7) = -3$: $\frac{f(0) - (-3)}{7} \le 2$ $f(0) + 3 \le 14$ $f(0) \le 11$

Step 3: Find an upper bound for $f(-1) + f(0)$.

• Why: The problem asks for the interval in which $f(-1) + f(0)$ lies. We have upper bounds for $f(-1)$ and $f(0)$, so we can find an upper bound for their sum.
• Adding the inequalities from Step 1 and Step 2: $f(-1) + f(0) \le 9 + 11$ $f(-1) + f(0) \le 20$

Step 4: Apply LMVT on the interval [-1, 0].

• Why: To find a lower bound for $f(-1) + f(0)$, we need to relate $f(-1)$ and $f(0)$. Applying LMVT on [-1, 0] will relate $f(0)$ and $f(-1)$ using $f'(c)$ for some $c$ in (-1, 0).
• Applying LMVT on the interval [-1, 0], there exists a $c_3 \in (-1, 0)$ such that: $f'(c_3) = \frac{f(0) - f(-1)}{0 - (-1)} = f(0) - f(-1)$
• Since $f'(x) \le 2$, we have $f'(c_3) \le 2$, so $f(0) - f(-1) \le 2$. This gives us $f(0) \le f(-1) + 2$.

Step 5: Find a lower bound for $f(-1)$.

• Why: We need a lower bound for $f(-1)$ to eventually find a lower bound for $f(-1) + f(0)$.
• From Step 1, $f'(c_1) = \frac{f(-1) - f(-7)}{6}$. We are given $f'(x) \le 2$, but we can't assume $f'(x)$ is always exactly 2. Let us consider the lower extreme value of $f'(x)$. Although we don't have a lower bound on $f'(x)$, we can establish a lower bound on $f(-1)$ by thinking about how "small" $f(-1)$ can be.

Step 6: Revisit LMVT on [-7, -1] and [-7, 0] to derive lower bounds for f(-1) and f(0).

• Why: The derivative is bounded above by 2. Let's use this information to derive the smallest possible values of f(-1) and f(0).
• For the interval [-7, -1]: $f'(c_1) = \frac{f(-1) - f(-7)}{-1 - (-7)} = \frac{f(-1) - (-3)}{6}$. Rearranging, $f(-1) = 6f'(c_1) - 3$.
• For the interval [-7, 0]: $f'(c_2) = \frac{f(0) - f(-7)}{0 - (-7)} = \frac{f(0) - (-3)}{7}$. Rearranging, $f(0) = 7f'(c_2) - 3$.
• Therefore, $f(-1) + f(0) = 6f'(c_1) - 3 + 7f'(c_2) - 3 = 6f'(c_1) + 7f'(c_2) - 6$.
• Since we do not have a lower bound on $f'(x)$, we can't directly find a lower bound on the sum $f(-1) + f(0)$ this way.

Step 7: Consider a specific function to find a lower bound.

• Why: Since we are looking for the interval in which all such functions' values lie, we can examine a specific function that satisfies the given conditions to help determine a lower bound.
• Consider the function $f(x) = 2x + 11$. Then $f(-7) = 2(-7) + 11 = -14 + 11 = -3$, and $f'(x) = 2 \le 2$. This function satisfies the given conditions.
• For this function, $f(-1) = 2(-1) + 11 = 9$ and $f(0) = 2(0) + 11 = 11$. Then $f(-1) + f(0) = 9 + 11 = 20$. This doesn't help us find a lower bound.
• Consider the function $f(x) = -\frac{3}{7}x - 3$. Then $f(-7) = -\frac{3}{7}(-7) - 3 = 3-3=0$, so this doesn't work.
• Consider the function $f(x) = 2x+11$. Then $f(-7) = -14+11 = -3$. And $f'(x) = 2$. This function satisfies the conditions. Then $f(-1) = 2(-1)+11 = 9$ and $f(0) = 11$. Thus $f(-1) + f(0) = 20$.
• Consider the function $f(x) = -\frac{3}{7}x -3$. Then $f(-7)=3-3=0 \neq -3$. This doesn't work.

Step 8: Consider the constant function $f(x) = -3$.

• Why: A constant function has a derivative of 0, which satisfies the condition $f'(x) \le 2$. Since $f(-7) = -3$, we can try this simple function.
• If $f(x) = -3$, then $f(-1) = -3$ and $f(0) = -3$. Thus $f(-1) + f(0) = -6$.

Step 9: Conclude the range for $f(-1) + f(0)$.

• Why: We have found an upper bound of 20 and a value of -6 for $f(-1) + f(0)$ with a valid function. This suggests the interval [-6, 20].
• Combining the upper bound $f(-1) + f(0) \le 20$ and the specific case where $f(-1) + f(0) = -6$, we can conclude that the interval for $f(-1) + f(0)$ is $[-6, 20]$.

Common Mistakes & Tips

• Remember that LMVT only guarantees the existence of a point $c$, not its specific value.
• Don't assume $f'(x)$ is always equal to its upper bound. Consider cases where it is less than 2.
• When searching for bounds, consider simple functions that satisfy the given conditions, such as linear or constant functions.

Summary

We used Lagrange's Mean Value Theorem to establish an upper bound for $f(-1) + f(0)$. By considering the constant function $f(x) = -3$, we found that $f(-1) + f(0)$ can be as low as -6. Therefore, the interval in which $f(-1) + f(0)$ lies is $[-6, 20]$.

The final answer is \boxed{[-6, 20]}, which corresponds to option (A).