The problem provided asks to determine the validity of the statement: "There exists a value $c$ in the open interval $1 < x < 7$ such that $f'(c) = -6$."

Given the function values in the table, we need to apply the Mean Value Theorem (MVT), which states that if a function $f$ is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one $c$ in $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Here, the function values are:

• $f(1) = -3$
• $f(4) = -18$
• $f(7) = -39$

We can calculate the slope of the secant line between any two points on the interval $[1, 7]$ to see if it matches $-6$.

Step 1: Calculate the secant slopes for the interval $[1, 7]$

$\frac{f(7) - f(1)}{7 - 1} = \frac{-39 - (-3)}{7 - 1} = \frac{-39 + 3}{6} = \frac{-36}{6} = -6$

Conclusion:

The secant slope between $x = 1$ and $x = 7$ is exactly $-6$, and since the function is differentiable on the open interval $1 < x < 7$, by the Mean Value Theorem, there must exist a value $c$ in $(1, 7)$ where $f'(c) = -6$.

Thus, the statement is true.

Would you like a further breakdown of any of the steps? Feel free to ask.

Related Questions:

1. How do you apply the Mean Value Theorem to other functions?
2. What does the concept of differentiability imply for a function?
3. How do you determine if a function is continuous over an interval?
4. What is the geometric interpretation of the Mean Value Theorem?
5. How do you solve for a point $c$ in the Mean Value Theorem problem?

Tip: Always check the conditions for the Mean Value Theorem: continuity on the closed interval and differentiability on the open interval!