Step 1: Understanding the Concept:
Lagrange's Mean Value Theorem (LMVT) guarantees that for a continuous and differentiable function on an interval $[a, b]$, there is at least one point $c$ in the open interval $(a, b)$ where the instantaneous rate of change (derivative) equals the average rate of change over the interval. We must calculate both sides of the theorem's equation and solve for $c$, ensuring the result lies within $(0, 1/2)$.
Step 2: Key Formula or Approach:
1. Expand the polynomial $f(x)$ for easier differentiation.
2. Find the derivative $f'(x)$ and formulate $f'(c)$.
3. Calculate the average rate of change: $\frac{f(b)-f(a)}{b-a}$.
4. Equate them, solve the resulting quadratic equation for $c$, and select the root satisfying $a<c<b$.
Step 3: Detailed Explanation:
Given $f(x) = x(x - 1)(x - 2)$. Let's expand this:
\[ f(x) = x(x^2 - 3x + 2) = x^3 - 3x^2 + 2x \]
Find the derivative $f'(x)$:
\[ f'(x) = 3x^2 - 6x + 2 \]
So, $f'(c) = 3c^2 - 6c + 2$.
Now, evaluate the function at the endpoints $a=0$ and $b=1/2$:
\[ f(a) = f(0) = 0(0 - 1)(0 - 2) = 0 \]
\[ f(b) = f(1/2) = (1/2)(1/2 - 1)(1/2 - 2) = (1/2)(-1/2)(-3/2) = 3/8 \]
Calculate the slope of the secant line:
\[ \frac{f(b) - f(a)}{b - a} = \frac{3/8 - 0}{1/2 - 0} = \frac{3/8}{1/2} = \frac{3}{8} \times 2 = \frac{3}{4} \]
Apply LMVT by equating the instantaneous slope to the secant slope:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
\[ 3c^2 - 6c + 2 = \frac{3}{4} \]
Clear the fraction by multiplying by 4:
\[ 12c^2 - 24c + 8 = 3 \]
\[ 12c^2 - 24c + 5 = 0 \]
Solve this quadratic equation using the quadratic formula $c = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:
\[ c = \frac{24 \pm \sqrt{(-24)^2 - 4(12)(5)}}{2(12)} \]
\[ c = \frac{24 \pm \sqrt{576 - 240}}{24} = \frac{24 \pm \sqrt{336}}{24} \]
Simplify the radical ($\sqrt{336} = \sqrt{16 \times 21} = 4\sqrt{21}$):
\[ c = \frac{24 \pm 4\sqrt{21}}{24} = 1 \pm \frac{\sqrt{21}}{6} \]
We must choose the value of $c$ that lies strictly within the interval $(0, 1/2)$.
We know $\sqrt{21}$ is between $\sqrt{16}=4$ and $\sqrt{25}=5$ (approx 4.58).
So $\frac{\sqrt{21}}{6}$ is approx $4.58/6 \approx 0.76$.
$c_1 = 1 + 0.76 = 1.76$, which is outside $(0, 0.5)$.
$c_2 = 1 - 0.76 = 0.24$, which is nicely inside the interval $(0, 0.5)$.
Therefore, the valid root is $1 - \frac{\sqrt{21}}{6}$.
Step 4: Final Answer:
The value of $c$ is $1 - \frac{\sqrt{21}}{6}$.