The question is about applying the Mean Value Theorem (MVT) to the polynomial $f(x) = 5x^2 - x$ over the interval $[0, 3]$. Let me break this down and solve for the value of $c$ satisfying the theorem.

Mean Value Theorem Formula

The Mean Value Theorem states that if $f(x)$ is continuous and differentiable on the interval $[a, b]$, then there exists a $c \in (a, b)$ such that:

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

Step 1: Calculate $f'(x)$ (the derivative of $f(x)$)

$f(x) = 5x^2 - x$ $f'(x) = 10x - 1$

Step 2: Evaluate $f(a)$ and $f(b)$ for $a = 0$ and $b = 3$

$f(0) = 5(0)^2 - 0 = 0$ $f(3) = 5(3)^2 - 3 = 45 - 3 = 42$

Step 3: Calculate the slope of the secant line

$\text{slope} = \frac{f(3) - f(0)}{3 - 0} = \frac{42 - 0}{3} = 14$

Step 4: Solve $f'(c) = 14$

$f'(c) = 10c - 1$ Set $f'(c) = 14$: $10c - 1 = 14$ $10c = 15$ $c = 1.5$

Final Answer:

The value of $c$ satisfying the Mean Value Theorem is: $c = 1.5$

Would you like further clarification or additional examples? Here are five follow-up questions for deeper understanding:

1. What are the necessary conditions for applying the Mean Value Theorem?
2. How is the derivative $f'(c)$ interpreted geometrically in the context of the MVT?
3. Can the MVT be applied to functions with discontinuities or sharp corners? Why or why not?
4. How would the solution change if the interval was altered, e.g., to $[1, 4]$?
5. How does the Mean Value Theorem relate to the Intermediate Value Theorem?

Tip: Always check the differentiability and continuity of the function before applying the Mean Value Theorem to avoid invalid assumptions.