1. \(f(x)\) is continuous on the closed interval \([a, b]\).

2. \(f(x)\) is differentiable on the open interval \((a, b)\).

Then there exists at least one \(c\) in the open interval \((a, b)\) such that:

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

In other words, there is at least one point \(c\) in the interval \((a, b)\) where the instantaneous rate of change (given by the derivative \(f'(c)\)) is equal to the average rate of change of the function over the interval \([a, b]\).

2. Like Rolle's Theorem, the MVT emphasizes the importance of continuity and differentiability. If the function is not continuous on the closed interval \([a, b]\) or not differentiable on the open interval \((a, b)\), the theorem does not apply.

3. The MVT is often used to approximate the value of \(f'(c)\) or to prove other theorems and results in calculus, including the First Mean Value Theorem for Integrals and the Second Mean Value Theorem.

4. The MVT has practical applications in physics and engineering, where it is used to analyze rates of change, average speeds, and other real-world problems.

5. While the MVT guarantees the existence of at least one \(c\) where \(f'(c)\) equals the average rate of change, it doesn't provide information about how many such points there might be. Multiple points with this property can exist within the interval.

Let \(f(x) = x^2 - 4x + 3\), and consider the closed interval \([1, 4]\).

a. Verify whether the conditions of the Mean Value Theorem (MVT) are satisfied for this function on the interval \([1, 4]\). State your reasons for each condition.

b. If the conditions of the MVT are satisfied, find at least one point \(c\) in the open interval \((1, 4)\) that satisfies the conclusion of the MVT. If the conditions are not satisfied, explain why.

a. Conditions of the Mean Value Theorem:

- Continuity on \([1, 4]\):** The function \(f(x) = x^2 - 4x + 3\) is a polynomial function, and polynomial functions are continuous everywhere. So, it is continuous on \([1, 4]\).

- Differentiability on \((1, 4)\):** The function is a polynomial, so it is differentiable everywhere. Thus, it is differentiable on \((1, 4)\).

Therefore, both conditions of the Mean Value Theorem are satisfied for this function on \([1, 4]\).

b. Since the conditions of the Mean Value Theorem are satisfied, we can conclude that there exists at least one point \(c\) in the open interval \((1, 4)\) such that:

\[f'(c) = \frac{f(4) - f(1)}{4 - 1}.\]

First, let's find the derivative of \(f(x)\):

\[f(x) = x^2 - 4x + 3\]

\[f'(x) = 2x - 4\]

Now, we set up the equation with the Mean Value Theorem:

\[2c - 4 = \frac{f(4) - f(1)}{4 - 1}\]

\[2c - 4 = \frac{(4^2 - 4(4) + 3) - (1^2 - 4(1) + 3)}{3}\]

\[2c - 4 = \frac{16 - 16 + 3 - 1 + 4 - 3}{3}\]

\[2c - 4 = \frac{3}{3}\]

\[2c - 4 = 1\]

Solving for \(c\):

\[2c = 5\]

\[c = \frac{5}{2}\]

Therefore, there is at least one point \(c = \frac{5}{2}\) in the open interval \((1, 4)\) that satisfies the conclusion of the Mean Value Theorem.

So, the point \(\left(\frac{5}{2}, f\left(\frac{5}{2}\right)\right)\) on the graph of \(f(x)\) has a tangent line that is parallel to the secant line connecting the points \((1, f(1))\) and \((4, f(4))\).

**Notes**-
**NCERT Solutions & Assignments**- NCERT Solution for Continuity And differentiability Exercise 5.1
- NCERT Solution for Continuity And differentiability Exercise 5.2
- NCERT Solution for Continuity And differentiability Exercise 5.3
- NCERT Solution for Continuity And differentiability Exercise 5.4
- Continuity and Differentiability class 12 Important questions

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