Math Problem Statement

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

Solution

To verify Rolle's Theorem for the given function f(x)=(x1)(x2)(x3)f(x) = (x - 1)(x - 2)(x - 3) on the interval [1,3][1, 3], we need to ensure the following conditions are satisfied:

Conditions of Rolle's Theorem:

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

Step 1: Check Continuity and Differentiability

The function f(x)=(x1)(x2)(x3)f(x) = (x - 1)(x - 2)(x - 3) is a polynomial. Polynomials are continuous and differentiable everywhere. Thus, f(x)f(x) is continuous on [1,3][1, 3] and differentiable on (1,3)(1, 3).

Step 2: Verify f(1)=f(3)f(1) = f(3)

f(1)=(11)(12)(13)=0f(1) = (1 - 1)(1 - 2)(1 - 3) = 0 f(3)=(31)(32)(33)=0f(3) = (3 - 1)(3 - 2)(3 - 3) = 0 Since f(1)=f(3)=0f(1) = f(3) = 0, the third condition is also satisfied.

Step 3: Find cc where f(c)=0f'(c) = 0

The derivative of f(x)f(x) is: f(x)=ddx[(x1)(x2)(x3)]f'(x) = \frac{d}{dx} \left[ (x - 1)(x - 2)(x - 3) \right] Using the product rule: f(x)=(x2)(x3)+(x1)(x3)+(x1)(x2)f'(x) = (x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2) Simplify: f(x)=(x2)(x3)+(x1)(x3)+(x1)(x2)f'(x) = (x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2) f(x)=(x25x+6)+(x24x+3)+(x23x+2)f'(x) = (x^2 - 5x + 6) + (x^2 - 4x + 3) + (x^2 - 3x + 2) f(x)=3x212x+11f'(x) = 3x^2 - 12x + 11

Now, solve f(x)=0f'(x) = 0 for xx: 3x212x+11=03x^2 - 12x + 11 = 0 Divide through by 3: x24x+113=0x^2 - 4x + \frac{11}{3} = 0 Solve using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here, a=1a = 1, b=4b = -4, and c=113c = \frac{11}{3}: x=(4)±(4)24(1)(113)2(1)x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)\left(\frac{11}{3}\right)}}{2(1)} x=4±164432x = \frac{4 \pm \sqrt{16 - \frac{44}{3}}}{2} x=4±4834432x = \frac{4 \pm \sqrt{\frac{48}{3} - \frac{44}{3}}}{2} x=4±432x = \frac{4 \pm \sqrt{\frac{4}{3}}}{2} x=4±232x = \frac{4 \pm \frac{2}{\sqrt{3}}}{2} x=2±13x = 2 \pm \frac{1}{\sqrt{3}} Rationalize: x=2±33x = 2 \pm \frac{\sqrt{3}}{3}

Thus, the critical points are: x=2+33andx=233.x = 2 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 2 - \frac{\sqrt{3}}{3}.

Step 4: Check if xx lies in (1,3)(1, 3)

Both x=2+33x = 2 + \frac{\sqrt{3}}{3} and x=233x = 2 - \frac{\sqrt{3}}{3} lie within the interval (1,3)(1, 3).

Conclusion

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

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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)=0f'(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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Polynomial Functions
Derivatives
Quadratic Equations

Formulas

Rolle's Theorem
Quadratic Formula
Product Rule for Differentiation

Theorems

Rolle's Theorem

Suitable Grade Level

Grades 11-12 (High School Calculus)