Verify the truth of Rolle's Theorem for the following functions :
`f(x)=x^(2)-4x+3" on "[1,3]`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = x^2 - 4x + 3 \) on the interval \([1, 3]\), we will follow these steps:

### Step 1: Check Continuity on the Closed Interval
Rolle's Theorem states that the function must be continuous on the closed interval \([a, b]\). Since \( f(x) \) is a polynomial function, it is continuous for all \( x \).

\[ \text{Conclusion: } f(x) \text{ is continuous on } [1, 3]. \]

### Step 2: Check Differentiability on the Open Interval
Next, we need to check if \( f(x) \) is differentiable on the open interval \( (1, 3) \). Again, since \( f(x) \) is a polynomial, it is differentiable everywhere.

\[ \text{Conclusion: } f(x) \text{ is differentiable on } (1, 3). \]

### Step 3: Check if \( f(a) = f(b) \)
Now we need to evaluate \( f(1) \) and \( f(3) \) to check if they are equal.

\[ f(1) = 1^2 - 4 \cdot 1 + 3 = 1 - 4 + 3 = 0 \]
\[ f(3) = 3^2 - 4 \cdot 3 + 3 = 9 - 12 + 3 = 0 \]

Since \( f(1) = f(3) = 0 \), we have satisfied this condition.

\[ \text{Conclusion: } f(1) = f(3). \]

### Step 4: Find \( c \) such that \( f'(c) = 0 \)
According to Rolle's Theorem, there exists at least one \( c \) in \( (1, 3) \) such that \( f'(c) = 0 \). First, we find the derivative of \( f(x) \):

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

Now we set the derivative equal to zero:

\[ 2x - 4 = 0 \implies 2x = 4 \implies x = 2. \]

Since \( 2 \) is in the open interval \( (1, 3) \), we have found our \( c \).

\[ \text{Conclusion: } c = 2 \text{ satisfies } f'(c) = 0. \]

### Final Conclusion
All conditions of Rolle's Theorem are satisfied:

1. \( f(x) \) is continuous on \([1, 3]\).
2. \( f(x) \) is differentiable on \( (1, 3) \).
3. \( f(1) = f(3) \).

Thus, we can conclude that Rolle's Theorem is verified for the function \( f(x) = x^2 - 4x + 3 \) on the interval \([1, 3]\).

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