Verify Rolle's theorem for f(x) = (x - 1)(x - 2)(x - 3) in [1.3].

To verify Rolle's Theorem for the given function \( f(x) = (x - 1)(x - 2)(x - 3) \) on the interval \([1, 3]\), we need to ensure the following conditions are satisfied:

### Conditions of Rolle's Theorem:
1. \( f(x) \) is continuous on \([1, 3]\).
2. \( f(x) \) is differentiable on \((1, 3)\).
3. \( f(1) = f(3) \).

### Step 1: Check Continuity and Differentiability
The function \( f(x) = (x - 1)(x - 2)(x - 3) \) is a polynomial. Polynomials are continuous and differentiable everywhere. Thus, \( f(x) \) is continuous on \([1, 3]\) and differentiable on \((1, 3)\).

### Step 2: Verify \( f(1) = f(3) \)
\[
f(1) = (1 - 1)(1 - 2)(1 - 3) = 0
\]
\[
f(3) = (3 - 1)(3 - 2)(3 - 3) = 0
\]
Since \( f(1) = f(3) = 0 \), the third condition is also satisfied.

### Step 3: Find \( c \) where \( f'(c) = 0 \)
The derivative of \( f(x) \) is:
\[
f'(x) = \frac{d}{dx} \left[ (x - 1)(x - 2)(x - 3) \right]
\]
Using the product rule:
\[
f'(x) = (x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2)
\]
Simplify:
\[
f'(x) = (x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2)
\]
\[
f'(x) = (x^2 - 5x + 6) + (x^2 - 4x + 3) + (x^2 - 3x + 2)
\]
\[
f'(x) = 3x^2 - 12x + 11
\]

Now, solve \( f'(x) = 0 \) for \( x \):
\[
3x^2 - 12x + 11 = 0
\]
Divide through by 3:
\[
x^2 - 4x + \frac{11}{3} = 0
\]
Solve using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -4 \), and \( c = \frac{11}{3} \):
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)\left(\frac{11}{3}\right)}}{2(1)}
\]
\[
x = \frac{4 \pm \sqrt{16 - \frac{44}{3}}}{2}
\]
\[
x = \frac{4 \pm \sqrt{\frac{48}{3} - \frac{44}{3}}}{2}
\]
\[
x = \frac{4 \pm \sqrt{\frac{4}{3}}}{2}
\]
\[
x = \frac{4 \pm \frac{2}{\sqrt{3}}}{2}
\]
\[
x = 2 \pm \frac{1}{\sqrt{3}}
\]
Rationalize:
\[
x = 2 \pm \frac{\sqrt{3}}{3}
\]

Thus, the critical points are:
\[
x = 2 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 2 - \frac{\sqrt{3}}{3}.
\]

### Step 4: Check if \( x \) lies in \((1, 3)\)
Both \( x = 2 + \frac{\sqrt{3}}{3} \) and \( x = 2 - \frac{\sqrt{3}}{3} \) lie within the interval \((1, 3)\).

### Conclusion
Rolle's Theorem is verified, as there exist at least two points \( c \) in \((1, 3)\) where \( f'(c) = 0 \).

Would you like further clarifications or visualizations? 

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**Relative Questions:**
1. What is the geometric interpretation of Rolle's Theorem in this context?
2. How does continuity and differentiability impact the application of Rolle's Theorem?
3. Can Rolle's Theorem have multiple points where \( f'(c) = 0 \)?
4. How would the analysis change if the function was not a polynomial?
5. What are real-life examples where Rolle's Theorem can be applied?

**Tip:** When solving polynomial derivatives, double-check for algebraic simplifications to ensure clarity and avoid errors.

Verify Rolle's theorem for f(x) = (x - 1)(x - 2)(x - 3) in [1,3].

Learn how to verify Rolle's Theorem for the polynomial function f(x) = (x - 1)(x - 2)(x - 3) on the interval [1,3]. This detailed solution covers key calculus concepts, including continuity, differentiability, and solving quadratic equations using the quadratic formula. Suitable for high school calculus students.

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