The value of the function `(x-1)(x-2)^2` at its maxima is

Options

A. 1
B. 2
C. 0
D. `(4)/(27)`

Explanation

To find the maximum value of the function \( f(x) = (x-1)(x-2)^2 \), we will follow these steps:

### Step 1: Expand the Function
First, we will expand the function to make differentiation easier.

\[ f(x) = (x-1)(x-2)^2 \]

Expanding \( (x-2)^2 \):

\[ (x-2)^2 = x^2 - 4x + 4 \]

Now substituting back into \( f(x) \):

\[ f(x) = (x-1)(x^2 - 4x + 4) \]

Now, distribute \( (x-1) \):

\[ f(x) = x(x^2 - 4x + 4) - 1(x^2 - 4x + 4) \]
\[ = x^3 - 4x^2 + 4x - x^2 + 4x - 4 \]
\[ = x^3 - 5x^2 + 8x - 4 \]

### Step 2: Differentiate the Function
Next, we differentiate \( f(x) \) with respect to \( x \):

\[ f'(x) = 3x^2 - 10x + 8 \]

### Step 3: Find Critical Points
To find the critical points, we set the derivative equal to zero:

\[ 3x^2 - 10x + 8 = 0 \]

Now we will factor or use the quadratic formula:

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 3, b = -10, c = 8 \):

\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} \]
\[ = \frac{10 \pm \sqrt{100 - 96}}{6} \]
\[ = \frac{10 \pm 2}{6} \]

This gives us:

\[ x = \frac{12}{6} = 2 \quad \text{and} \quad x = \frac{8}{6} = \frac{4}{3} \]

### Step 4: Determine Maxima or Minima
Now we will use the second derivative test to determine whether these critical points are maxima or minima.

First, we find the second derivative:

\[ f''(x) = 6x - 10 \]

Now we evaluate \( f''(x) \) at the critical points:

1. For \( x = \frac{4}{3} \):

\[ f''\left(\frac{4}{3}\right) = 6 \cdot \frac{4}{3} - 10 = 8 - 10 = -2 \quad (\text{negative, hence maximum}) \]

2. For \( x = 2 \):

\[ f''(2) = 6 \cdot 2 - 10 = 12 - 10 = 2 \quad (\text{positive, hence minimum}) \]

### Step 5: Calculate Maximum Value
The maximum value occurs at \( x = \frac{4}{3} \). Now we calculate \( f\left(\frac{4}{3}\right) \):

\[ f\left(\frac{4}{3}\right) = \left(\frac{4}{3} - 1\right)\left(\frac{4}{3} - 2\right)^2 \]
\[ = \left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)^2 \]
\[ = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27} \]

Thus, the maximum value of the function \( f(x) \) is:

\[ \boxed{\frac{4}{27}} \]