To solve this problem using the Mean Value Theorem (MVT), let’s proceed step-by-step:

Problem Statement Recap

You are asked to find all points $c$ in the interval $(0, 2)$ such that: $f(2) - f(0) = f'(c) \cdot (2 - 0)$ where $f(x) = x^3$.

Step 1: Apply the Mean Value Theorem

The Mean Value Theorem states that if $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a point $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$ In our case:

• $f(x) = x^3$, so it is continuous and differentiable everywhere on $[0, 2]$.
• We set $a = 0$ and $b = 2$.

Step 2: Calculate $f(2)$ and $f(0)$

For $f(x) = x^3$: $f(2) = 2^3 = 8$ $f(0) = 0^3 = 0$

Thus, $f(2) - f(0) = 8 - 0 = 8$

Step 3: Apply MVT to Find $c$

Using the MVT, we need $c \in (0, 2)$ such that: $f'(c) \cdot (2 - 0) = 8$ or equivalently, $f'(c) = \frac{8}{2} = 4$

Step 4: Calculate $f'(x)$ and Solve for $c$

Now, find $f'(x)$ for $f(x) = x^3$: $f'(x) = 3x^2$

Set $f'(c) = 4$: $3c^2 = 4$ $c^2 = \frac{4}{3}$ $c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm \frac{2 \sqrt{3}}{3}$

Since we need $c \in (0, 2)$, we take the positive solution: $c = \frac{2 \sqrt{3}}{3}$

Conclusion

The point $c$ in the interval $(0, 2)$ that satisfies the Mean Value Theorem for this function is: $c = \frac{2 \sqrt{3}}{3}$

Would you like further details or have any questions?

Here are some related questions to deepen your understanding:

1. How would the result change if $f(x)$ was a different polynomial function?
2. What are the conditions required for the Mean Value Theorem to hold?
3. Can you apply the Mean Value Theorem to non-polynomial functions?
4. How does the Mean Value Theorem relate to Rolle's Theorem?
5. What is the geometric interpretation of the Mean Value Theorem?

Tip: When using the Mean Value Theorem, ensure the function is continuous on the closed interval and differentiable on the open interval to apply it correctly.