The maximum value of the function f(x) given by
`f(x)=x(x-1)^2,0ltxlt2` , is

Options

A. 0
B. `4//27`
C. -4
D. `1//4`

Answer

To find the maximum value of the function \( f(x) = x(x-1)^2 \) in the interval \( 0 < x < 2 \), we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first need to differentiate the function \( f(x) \).

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

Using the product rule, where \( u = x \) and \( v = (x-1)^2 \):

\[ f'(x) = u'v + uv' \]

Calculating \( u' \) and \( v' \):

\[ u' = 1 \]
\[ v = (x-1)^2 \quad \Rightarrow \quad v' = 2(x-1) \]

Now applying the product rule:

\[ f'(x) = 1 \cdot (x-1)^2 + x \cdot 2(x-1) \]
\[ = (x-1)^2 + 2x(x-1) \]
\[ = (x-1)^2 + 2x^2 - 2x \]
\[ = (x-1)^2 + 2x^2 - 2x \]
\[ = x^2 - 2x + 1 + 2x^2 - 2x \]
\[ = 3x^2 - 4x + 1 \]

### Step 2: Set the derivative to zero
Now we set the derivative equal to zero to find the critical points:

\[ 3x^2 - 4x + 1 = 0 \]

### Step 3: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 3 \), \( b = -4 \), and \( c = 1 \).

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]
\[ = \frac{4 \pm \sqrt{16 - 12}}{6} \]
\[ = \frac{4 \pm \sqrt{4}}{6} \]
\[ = \frac{4 \pm 2}{6} \]

Calculating the two possible values:

1. \( x = \frac{6}{6} = 1 \)
2. \( x = \frac{2}{6} = \frac{1}{3} \)

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical points \( x = 1 \) and \( x = \frac{1}{3} \), as well as the endpoints of the interval \( x = 0 \) and \( x = 2 \).

1. \( f(0) = 0(0-1)^2 = 0 \)
2. \( f(1) = 1(1-1)^2 = 0 \)
3. \( f\left(\frac{1}{3}\right) = \frac{1}{3}\left(\frac{1}{3} - 1\right)^2 = \frac{1}{3}\left(-\frac{2}{3}\right)^2 = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27} \)
4. \( f(2) = 2(2-1)^2 = 2(1)^2 = 2 \)

### Step 5: Determine the maximum value
Now we compare the values:

- \( f(0) = 0 \)
- \( f(1) = 0 \)
- \( f\left(\frac{1}{3}\right) = \frac{4}{27} \)
- \( f(2) = 2 \)

The maximum value of \( f(x) \) in the interval \( 0 < x < 2 \) is:

\[ \text{Maximum value} = 2 \]

### Final Answer
The maximum value of the function \( f(x) = x(x-1)^2 \) for \( 0 < x < 2 \) is \( \boxed{2} \).