The Mean Value Theorem (MVT) states that if $f(x)$ 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:

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

Here, the function is $f(x) = x^2$, and the interval is $[2, 4]$.

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Step 1: Compute $f(b) - f(a)$ and $b - a$

Let $a = 2$ and $b = 4$.

$f(4) = 4^2 = 16, \quad f(2) = 2^2 = 4$

$f(4) - f(2) = 16 - 4 = 12, \quad b - a = 4 - 2 = 2$

$\frac{f(b) - f(a)}{b - a} = \frac{12}{2} = 6$

Thus, $f'(c) = 6$.

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Step 2: Differentiate $f(x)$ and solve for $c$

The derivative of $f(x) = x^2$ is:

$f'(x) = 2x$

Set $f'(c) = 6$:

$2c = 6 \implies c = 3$

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Step 3: Verify that $c \in (2, 4)$

The value $c = 3$ lies in the open interval $(2, 4)$. Therefore, it satisfies the Mean Value Theorem.

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Final Answer:

The value of $c$ that satisfies the Mean Value Theorem is:

$\boxed{c = 3}$

Do you want details or have any questions?

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Related Questions:

1. What are the conditions for the Mean Value Theorem to hold?
2. How does the Mean Value Theorem differ from Rolle's Theorem?
3. Can you apply the Mean Value Theorem to $f(x) = \sin(x)$ on $[0, \pi]$?
4. How do you compute the derivative of more complex functions for the MVT?
5. How would the MVT apply if $f(x) = |x|$ on $[-1, 1]$?

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Tip:

For quadratic functions like $f(x) = x^2$, the $c$ in the MVT often corresponds to the midpoint of the interval when symmetric behavior is present.