The problem is asking whether the Mean Value Theorem (MVT) can be applied to the function $f(x) = x^2$ on the closed interval $[-1, 7]$, and if so, to find all numbers $c$ in the open interval $(-1, 7)$ such that the derivative $f'(c)$ equals the average rate of change of the function on this interval.

Step 1: Conditions for applying the Mean Value Theorem (MVT)

The MVT can be applied if:

1. The function $f(x)$ is continuous on the closed interval $[-1, 7]$.
2. The function $f(x)$ is differentiable on the open interval $(-1, 7)$.

• The function $f(x) = x^2$ is a polynomial, which is continuous and differentiable everywhere on $\mathbb{R}$. Therefore, both conditions of the MVT are satisfied on the interval $[-1, 7]$.

Step 2: Apply the Mean Value Theorem

The MVT states that if the conditions are met, there exists at least one number $c$ in the open interval $(-1, 7)$ such that: $f'(c) = \frac{f(7) - f(-1)}{7 - (-1)}.$ Now, we will calculate each part:

• $f(7) = 7^2 = 49$
• $f(-1) = (-1)^2 = 1$
• The denominator is $7 - (-1) = 7 + 1 = 8$

So the right-hand side of the equation is: $\frac{f(7) - f(-1)}{7 - (-1)} = \frac{49 - 1}{8} = \frac{48}{8} = 6.$

Step 3: Find $c$

We need to find $c$ such that: $f'(c) = 6.$ First, find the derivative of $f(x) = x^2$: $f'(x) = 2x.$ Set this equal to 6: $2c = 6.$ Solving for $c$: $c = 3.$

Conclusion:

The Mean Value Theorem can be applied to the function $f(x) = x^2$ on the interval $[-1, 7]$, and the value of $c$ in the open interval $(-1, 7)$ is $c = 3$.

Would you like more details on any of these steps or related concepts?

Here are 5 related questions for further exploration:

1. Can the Mean Value Theorem be applied to a function with a discontinuity?
2. What happens if a function is continuous but not differentiable on an interval?
3. How does the Mean Value Theorem relate to Rolle's Theorem?
4. How would the result change if the interval was different, such as $[1, 5]$?
5. Could there be more than one value of $c$ for certain functions?

Tip: The Mean Value Theorem guarantees at least one point $c$, but for more complex functions, there could be multiple points satisfying the condition.