Examine the validity and conclusion of Lagrange's Mean Value theorem for the function f(x) = x(x - 1)(x - 2) for every x in [0, 1/2].

Steps to Solve

1. Check if the function is continuous on the interval The function ( f(x) = x(x - 1)(x - 2) ) is a polynomial, and all polynomials are continuous everywhere. Thus, it is continuous on the closed interval ([0, \frac{1}{2}]).

2. Check if the function is differentiable on the interval Being a polynomial, ( f(x) ) is also differentiable everywhere. Therefore, it is differentiable on the open interval ((0, \frac{1}{2})).

3. Calculate the function values at the endpoints We determine ( f(0) ) and ( f(\frac{1}{2}) ): [ f(0) = 0(0 - 1)(0 - 2) = 0 ] [ f\left(\frac{1}{2}\right) = \frac{1}{2}\left(\frac{1}{2} - 1\right)\left(\frac{1}{2} - 2\right) = \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right) = \frac{3}{8} ]

4. Apply Lagrange's Mean Value Theorem (MVT) According to MVT, there exists at least one ( c ) in ( (0, \frac{1}{2}) ) such that: [ f'(c) = \frac{f(b) - f(a)}{b - a} ] Here, ( a = 0 ) and ( b = \frac{1}{2} ). So we calculate: [ \frac{f\left(\frac{1}{2}\right) - f(0)}{\frac{1}{2} - 0} = \frac{\frac{3}{8} - 0}{\frac{1}{2}} = \frac{3}{8} \cdot \frac{2}{1} = \frac{3}{4} ]

5. Find the derivative of the function Next, we compute ( f'(x) ): [ f(x) = x^3 - 3x^2 + 2x ] Using basic derivatives: [ f'(x) = 3x^2 - 6x + 2 ]

6. Set the derivative equal to the average slope We set ( f'(c) = \frac{3}{4} ): [ 3c^2 - 6c + 2 = \frac{3}{4} ] Multiply through by 4 to eliminate the fraction: [ 12c^2 - 24c + 8 = 3 ] Rearranging gives: [ 12c^2 - 24c + 5 = 0 ]

7. Solve the quadratic equation Using the quadratic formula ( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ): Here, ( a = 12, b = -24, c = 5 ): [ c = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 12 \cdot 5}}{2 \cdot 12} ] Calculating the discriminant: [ 576 - 240 = 336 ] So, [ c = \frac{24 \pm \sqrt{336}}{24} = \frac{24 \pm 4\sqrt{21}}{24} = 1 \pm \frac{\sqrt{21}}{6} ]

8. Determine the value of ( c ) within ( (0, \frac{1}{2}) ) Estimate ( \sqrt{21} \approx 4.58 ): Thus, ( 1 - \frac{4.58}{6} ) is approximately ( 0.23 ), which lies within ((0, \frac{1}{2})).