The Mean Value Theorem (MVT) 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 $c \in (a, b)$ such that:

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

Step 1: Verify if the MVT applies

To apply the MVT, we need to check if the function $f(x) = 3x^2 - x$ meets the conditions of the theorem:

• Continuity: Since $f(x) = 3x^2 - x$ is a polynomial, it is continuous everywhere, so it is continuous on $[-1, 2]$.
• Differentiability: Polynomials are differentiable everywhere, so $f(x)$ is differentiable on $(-1, 2)$.

Since both conditions are satisfied, the MVT applies.

Step 2: Apply the Mean Value Theorem

Next, we need to find $c \in (-1, 2)$ such that:

$f'(c) = \frac{f(2) - f(-1)}{2 - (-1)}$

2.1 Calculate $f(b) - f(a)$

• $f(2) = 3(2)^2 - 2 = 12 - 2 = 10$
• $f(-1) = 3(-1)^2 - (-1) = 3(1) + 1 = 4$

Now, calculate the difference:

$f(2) - f(-1) = 10 - 4 = 6$

2.2 Compute the slope of the secant line

$\frac{f(2) - f(-1)}{2 - (-1)} = \frac{6}{3} = 2$

Thus, we need to find $c \in (-1, 2)$ such that:

$f'(c) = 2$

Step 3: Find $f'(x)$

The derivative of $f(x) = 3x^2 - x$ is:

$f'(x) = 6x - 1$

We want $f'(c) = 2$, so solve:

$6c - 1 = 2$

$6c = 3 \quad \Rightarrow \quad c = \frac{1}{2}$

Conclusion:

The value of $c$ guaranteed by the MVT is $c = \frac{1}{2}$.

Do you have any questions or would you like more details?

Here are 5 related questions to expand on this topic:

1. Can the MVT apply to non-polynomial functions? If so, under what conditions?
2. How does the Rolle's Theorem relate to the MVT?
3. Can there be more than one value of $c$ guaranteed by the MVT for certain functions?
4. What is the geometric interpretation of the MVT?
5. How do you verify differentiability for more complex functions?

Tip: When using the MVT, always check both continuity and differentiability on the appropriate intervals before proceeding with the calculation.