Understanding the Mean Value Theorem

Oct 13, 2025 by thesweetdesigns.com 71 views

**Let $f$ be the function given by $f(x) = x^3$. What are all values of $c$ that satisfy the conclusion of the Mean Value Theorem on the closed interval $[-1, 2]$?**

Understanding the Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus that states that for a function $f$ that is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , there exists a point $c$ in $(a, b)$ such that the derivative of the function at $c$ is equal to the average rate of change of the function over the interval $[a, b]$ .

The Formula for the Mean Value Theorem

The formula for the MVT is given by:

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

where $f'(c)$ is the derivative of the function $f$ at the point $c$ , and $\frac{f(b) - f(a)}{b - a}$ is the average rate of change of the function over the interval $[a, b]$ .

Applying the Mean Value Theorem to the Given Function

In this problem, we are given the function $f(x) = x^3$ and the closed interval $[-1, 2]$ . To apply the MVT, we need to find the derivative of the function $f$ and then use the formula for the MVT.

Finding the Derivative of the Function

The derivative of the function $f(x) = x^3$ is given by:

f'(x) = 3x^2

Using the Formula for the Mean Value Theorem

Now, we can use the formula for the MVT to find the values of $c$ that satisfy the conclusion of the MVT on the closed interval $[-1, 2]$ .

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

f'(c) = \frac{8 - (-1)}{3}

f'(c) = \frac{9}{3}

f'(c) = 3

Solving for $c$

Now, we need to solve for $c$ by setting the derivative of the function equal to the average rate of change of the function over the interval $[-1, 2]$ .

3c^2 = 3

c^2 = 1

c = \pm 1

Conclusion

Therefore, the values of $c$ that satisfy the conclusion of the Mean Value Theorem on the closed interval $[-1, 2]$ are $c = 1$ and $c = -1$ .

Q&A

Q: What is the Mean Value Theorem?

A: The Mean Value Theorem is a fundamental concept in calculus that states that for a function $f$ that is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , there exists a point $c$ in $(a, b)$ such that the derivative of the function at $c$ is equal to the average rate of change of the function over the interval $[a, b]$ .

Q: What is the formula for the Mean Value Theorem?

A: The formula for the MVT is given by:

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

Q: How do I apply the Mean Value Theorem to a given function?

A: To apply the MVT, you need to find the derivative of the function and then use the formula for the MVT.

Q: What is the derivative of the function $f(x) = x^3$ ?

A: The derivative of the function $f(x) = x^3$ is given by:

f'(x) = 3x^2

Q: How do I find the values of $c$ that satisfy the conclusion of the Mean Value Theorem?

A: To find the values of $c$ that satisfy the conclusion of the MVT, you need to set the derivative of the function equal to the average rate of change of the function over the interval $[a, b]$ and solve for $c$ .

Q: What are the values of $c$ that satisfy the conclusion of the Mean Value Theorem on the closed interval $[-1, 2]$ ?

A: The values of $c$ that satisfy the conclusion of the Mean Value Theorem on the closed interval $[-1, 2]$ are $c = 1$ and $c = -1$ .