Verify the truth of Rolle's Theorem for the following functions :
`f(x)=x^(2)-5x+4" on "[1,4]`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = x^2 - 5x + 4 \) on the interval \([1, 4]\), we need to check the three conditions of Rolle's Theorem:

1. **Continuity on the closed interval \([a, b]\)**.
2. **Differentiability on the open interval \((a, b)\)**.
3. **Equality of function values at the endpoints, i.e., \( f(a) = f(b) \)**.

Let's go through these conditions step by step.

### Step 1: Check Continuity

The function \( f(x) = x^2 - 5x + 4 \) is a polynomial function. Polynomial functions are continuous everywhere, including the closed interval \([1, 4]\).

**Conclusion**: \( f(x) \) is continuous on \([1, 4]\).

### Step 2: Check Differentiability

Since \( f(x) \) is a polynomial function, it is also differentiable everywhere, including the open interval \((1, 4)\).

**Conclusion**: \( f(x) \) is differentiable on \((1, 4)\).

### Step 3: Check Function Values at Endpoints

Now we need to evaluate \( f(1) \) and \( f(4) \):

- Calculate \( f(1) \):
\[ f(1) = 1^2 - 5 \cdot 1 + 4 = 1 - 5 + 4 = 0 \]

- Calculate \( f(4) \):
\[ f(4) = 4^2 - 5 \cdot 4 + 4 = 16 - 20 + 4 = 0 \]

Since \( f(1) = 0 \) and \( f(4) = 0 \), we have \( f(1) = f(4) \).

**Conclusion**: \( f(1) = f(4) \).

### Final Conclusion

Since all three conditions of Rolle's Theorem are satisfied, we can conclude that there exists at least one point \( c \) in the open interval \((1, 4)\) such that \( f'(c) = 0 \).

### Finding the Value of \( c \)

To find \( c \), we first need to compute the derivative \( f'(x) \):
\[ f'(x) = 2x - 5 \]
Now, we set the derivative equal to zero to find \( c \):
\[ 2c - 5 = 0 \implies 2c = 5 \implies c = \frac{5}{2} = 2.5 \]

Thus, there exists a point \( c = 2.5 \) in the interval \((1, 4)\) such that \( f'(c) = 0 \).

### Summary

We have verified that Rolle's Theorem holds for the function \( f(x) = x^2 - 5x + 4 \) on the interval \([1, 4]\), and we found \( c = 2.5 \) where \( f'(c) = 0 \).

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