To solve this problem, we'll apply the Mean Value Theorem (MVT). According to the MVT, for a function $f(x)$ that is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists at least one point $c \in (a, b)$ such that:

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

Given Information:

• Function: $f(x) = 4x^2 - 4x - 3$
• Interval: $[-1, 0]$

Step 1: Calculate $f(a)$ and $f(b)$.

Here, $a = -1$ and $b = 0$.

$f(-1) = 4(-1)^2 - 4(-1) - 3 = 4 + 4 - 3 = 5.$

$f(0) = 4(0)^2 - 4(0) - 3 = -3.$

Step 2: Compute the average rate of change.

$\frac{f(b) - f(a)}{b - a} = \frac{f(0) - f(-1)}{0 - (-1)} = \frac{-3 - 5}{1} = -8.$

Step 3: Find $f'(x)$.

Differentiate $f(x)$: $f'(x) = \frac{d}{dx}(4x^2 - 4x - 3) = 8x - 4.$

Step 4: Set $f'(c) = -8$ and solve for $c$.

$8c - 4 = -8.$

$8c = -8 + 4 = -4.$

$c = -\frac{1}{2}.$

Final Answer:

The value of $c$ that satisfies the Mean Value Theorem for the given function and interval is $c = -\frac{1}{2}$.

Would you like a deeper explanation of any of these steps, or do you have any questions?

Related Questions:

1. What are the conditions for applying the Mean Value Theorem?
2. How do you find the derivative of a quadratic function?
3. Can the Mean Value Theorem be applied if the function is not differentiable on the interval?
4. What is the geometric interpretation of the Mean Value Theorem?
5. How does the Mean Value Theorem relate to Rolle's Theorem?

Tip:

Always verify that the function is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ before applying the Mean Value Theorem.