To solve this problem using the Mean Value Theorem (MVT), we need to follow these steps:

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem states that 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 number $c$ in the interval $(a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$ This equation implies that the derivative at some point $c$ in $(a, b)$ is equal to the average rate of change of the function over the interval $[a, b]$.

Step 2: Define the Function and Derivative

Given: $f(x) = x^3 + 2x^2 - x + 4$ The derivative of $f(x)$ is: $f'(x) = 3x^2 + 4x - 1$

Step 3: Calculate the Average Rate of Change

The average rate of change of $f(x)$ over the interval $[-1, 3]$ is given by: $\frac{f(3) - f(-1)}{3 - (-1)}$

First, compute $f(3)$ and $f(-1)$:

$f(3) = 3^3 + 2(3)^2 - 3 + 4 = 27 + 18 - 3 + 4 = 46$ $f(-1) = (-1)^3 + 2(-1)^2 - (-1) + 4 = -1 + 2 + 1 + 4 = 6$

So, the average rate of change is: $\frac{46 - 6}{3 - (-1)} = \frac{40}{4} = 10$

Step 4: Set the Derivative Equal to the Average Rate of Change

According to the MVT, there exists some $c$ in the interval $(-1, 3)$ such that: $f'(c) = 10$ Substitute $f'(x) = 3x^2 + 4x - 1$ into the equation: $3x^2 + 4x - 1 = 10$

Step 5: Solve the Quadratic Equation

Simplify the equation: $3x^2 + 4x - 11 = 0$

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 3$, $b = 4$, and $c = -11$. Plugging in these values: $x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-11)}}{2(3)}$ $x = \frac{-4 \pm \sqrt{16 + 132}}{6}$ $x = \frac{-4 \pm \sqrt{148}}{6}$ $x = \frac{-4 \pm \sqrt{4 \times 37}}{6} = \frac{-4 \pm 2\sqrt{37}}{6} = \frac{-2 \pm \sqrt{37}}{3}$

Step 6: Find the Valid Solutions

Evaluate the two possible solutions: $x = \frac{-2 + \sqrt{37}}{3} \quad \text{and} \quad x = \frac{-2 - \sqrt{37}}{3}$

Check if these solutions are within the interval $(-1, 3)$:

• $\frac{-2 + \sqrt{37}}{3}$ is approximately $1.28$, which is within the interval $(-1, 3)$.
• $\frac{-2 - \sqrt{37}}{3}$ is approximately $-2.61$, which is not within the interval $(-1, 3)$.

Thus, the only valid solution is: $x = \frac{-2 + \sqrt{37}}{3}$

Final Answer:

The value of $x$ in the interval $(-1, 3)$ that satisfies the Mean Value Theorem is $\frac{-2 + \sqrt{37}}{3}$.

Would you like further details or have any questions about this process?

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Here are 5 related questions for further exploration:

1. What are the conditions for applying the Mean Value Theorem?
2. How can you interpret the result of the Mean Value Theorem geometrically?
3. What happens if the function is not differentiable at some point within the interval?
4. How does the Mean Value Theorem relate to Rolle's Theorem?
5. Can the Mean Value Theorem guarantee multiple values of $c$ in some cases?

Tip: When solving a quadratic equation, always consider whether both roots are within the required interval.