If C C C Is The Number That Satisfies The Conclusion Of The Mean Value Theorem For F ( X ) = X 3 − 2 X 2 F(x)=x^3-2x^2 F ( X ) = X 3 − 2 X 2 On The Interval 0 ≤ X ≤ 2 0 \leq X \leq 2 0 ≤ X ≤ 2 , Then C = C= C = A. 0 B. 1 / 2 1 / 2 1/2 C. 1 D. 4 / 3 4 / 3 4/3 E. 2

If $c$ is the number that satisfies the conclusion of the Mean Value Theorem for $f(x)=x^3-2x^2$ on the interval $0 \leq x \leq 2$, then $c=$

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Introduction

The Mean Value Theorem (MVT) is a fundamental concept in calculus that provides a powerful tool for analyzing the behavior of functions. It 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 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 $[a, b]$. In this article, we will apply the MVT to the function $f(x) = x^3 - 2x^2$ on the interval $0 \leq x \leq 2$ and find the value of $c$ that satisfies the conclusion of the MVT.

The Mean Value Theorem

The MVT is a theorem that provides a way to relate the derivative of a function to the average rate of change of the function over an interval. It is a powerful tool for analyzing the behavior of functions and is widely used in calculus and other areas of mathematics.

Theorem 1: (Mean Value Theorem) Let $f(x)$ be a function that is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. Then, there exists a point $c$ in $(a, b)$ such that

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

where $f'(c)$ is the derivative of the function at $c$.

Applying the Mean Value Theorem to $f(x) = x^3 - 2x^2$

To apply the MVT to the function $f(x) = x^3 - 2x^2$, we need to find the derivative of the function and evaluate it at the point $c$ that satisfies the conclusion of the MVT.

Step 1: Find the derivative of the function $f(x) = x^3 - 2x^2$.

Using the power rule for differentiation, we get

$$f'(x) = 3x^2 - 4x$$

Step 2: Evaluate the derivative at the point $c$ that satisfies the conclusion of the MVT.

We need to find the value of $c$ that satisfies the conclusion of the MVT, which is given by the equation

$$f'(c) = \frac{f(2) - f(0)}{2 - 0}$$

We can evaluate the derivative at the point $c$ by substituting the value of $c$ into the equation for the derivative.

Finding the Value of $c$

To find the value of $c$, we need to solve the equation

$$f'(c) = \frac{f(2) - f(0)}{2 - 0}$$

We can start by evaluating the function at the endpoints of the interval $0 \leq x \leq 2$.

Step 1: Evaluate the function at the endpoint $x = 0$.

We get

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

Step 2: Evaluate the function at the endpoint $x = 2$.

We get

$$f(2) = 2^3 - 2(2)^2 = 8 - 8 = 0$$

Step 3: Substitute the values of $f(0)$ and $f(2)$ into the equation for the derivative.

We get

$$f'(c) = \frac{f(2) - f(0)}{2 - 0} = \frac{0 - 0}{2 - 0} = 0$$

Step 4: Solve the equation $f'(c) = 0$.

We get

$$3c^2 - 4c = 0$$

Factoring out the common term $c$, we get

$$c(3c - 4) = 0$$

This gives us two possible values for $c$: $c = 0$ and $c = \frac{4}{3}$.

However, we need to check if these values of $c$ are in the interval $0 \leq x \leq 2$.

Step 5: Check if the values of $c$ are in the interval $0 \leq x \leq 2$.

We get

$$0 \leq c \leq 2$$

This means that the value of $c$ that satisfies the conclusion of the MVT is $c = \frac{4}{3}$.

Conclusion

In this article, we applied the Mean Value Theorem to the function $f(x) = x^3 - 2x^2$ on the interval $0 \leq x \leq 2$ and found the value of $c$ that satisfies the conclusion of the MVT. We showed that the value of $c$ is $c = \frac{4}{3}$.

Answer: The correct answer is D. $\frac{4}{3}$.
Q&A: Mean Value Theorem and Its Applications

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Introduction

The Mean Value Theorem (MVT) is a fundamental concept in calculus that provides a powerful tool for analyzing the behavior of functions. In our previous article, we applied the MVT to the function $f(x) = x^3 - 2x^2$ on the interval $0 \leq x \leq 2$ and found the value of $c$ that satisfies the conclusion of the MVT. In this article, we will answer some frequently asked questions about the MVT and its applications.

Q1: What is the Mean Value Theorem?

A1: The Mean Value Theorem is a theorem that provides a way to relate the derivative of a function to the average rate of change of the function over an interval. It 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 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 $[a, b]$.

Q2: What are the conditions for the Mean Value Theorem to hold?

A2: The conditions for the Mean Value Theorem to hold are:

• The function $f(x)$ must be continuous on the closed interval $[a, b]$.
• The function $f(x)$ must be differentiable on the open interval $(a, b)$.

Q3: How do I apply the Mean Value Theorem to a function?

A3: To apply the Mean Value Theorem to a function, you need to follow these steps:

• Find the derivative of the function.
• Evaluate the function at the endpoints of the interval.
• Substitute the values of the function into the equation for the derivative.
• Solve the equation for the derivative.

Q4: What is the significance of the Mean Value Theorem?

A4: The Mean Value Theorem is significant because it provides a way to relate the derivative of a function to the average rate of change of the function over an interval. This is useful for analyzing the behavior of functions and for solving optimization problems.

Q5: Can the Mean Value Theorem be applied to any function?

A5: No, the Mean Value Theorem cannot be applied to any function. The function must be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.

Q6: What are some common applications of the Mean Value Theorem?

A6: Some common applications of the Mean Value Theorem include:

• Finding the maximum and minimum values of a function.
• Solving optimization problems.
• Analyzing the behavior of functions.

Q7: How do I use the Mean Value Theorem to find the maximum and minimum values of a function?

A7: To use the Mean Value Theorem to find the maximum and minimum values of a function, you need to follow these steps:

• Find the derivative of the function.
• Evaluate the function at the endpoints of the interval.
• Substitute the values of the function into the equation for the derivative.
• Solve the equation for the derivative.
• Use the critical points to find the maximum and minimum values of the function.

Q8: Can the Mean Value Theorem be used to find the absolute maximum and minimum values of a function?

A8: Yes, the Mean Value Theorem can be used to find the absolute maximum and minimum values of a function.

Q9: How do I use the Mean Value Theorem to analyze the behavior of a function?

A9: To use the Mean Value Theorem to analyze the behavior of a function, you need to follow these steps:

• Find the derivative of the function.
• Evaluate the function at the endpoints of the interval.
• Substitute the values of the function into the equation for the derivative.
• Solve the equation for the derivative.
• Use the critical points to analyze the behavior of the function.

Q10: What are some common mistakes to avoid when applying the Mean Value Theorem?

A10: Some common mistakes to avoid when applying the Mean Value Theorem include:

• Not checking if the function is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.
• Not evaluating the function at the endpoints of the interval.
• Not substituting the values of the function into the equation for the derivative.
• Not solving the equation for the derivative.

Conclusion

In this article, we answered some frequently asked questions about the Mean Value Theorem and its applications. We hope that this article has been helpful in understanding the Mean Value Theorem and its significance in calculus.