Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=(x-1)(x-2)^(2)`.

Answer

To determine the intervals in which the function \( f(x) = (x-1)(x-2)^2 \) is strictly increasing or strictly decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

Using the product rule, where \( u = (x-1) \) and \( v = (x-2)^2 \):
- The derivative of \( u \) is \( u' = 1 \).
- The derivative of \( v \) is \( v' = 2(x-2) \cdot 1 = 2(x-2) \).

Now, applying the product rule:
\[ f'(x) = u'v + uv' = (1)(x-2)^2 + (x-1)(2(x-2)) \]
\[ = (x-2)^2 + 2(x-1)(x-2) \]

### Step 2: Simplify the derivative
Now we simplify the expression:
\[ f'(x) = (x-2)^2 + 2(x-1)(x-2) \]
Expanding \( 2(x-1)(x-2) \):
\[ = (x-2)^2 + 2(x^2 - 3x + 2) \]
\[ = (x-2)^2 + 2x^2 - 6x + 4 \]
Now, \( (x-2)^2 = x^2 - 4x + 4 \), so:
\[ f'(x) = x^2 - 4x + 4 + 2x^2 - 6x + 4 \]
\[ = 3x^2 - 10x + 8 \]

### Step 3: Find critical points
To find the critical points, we set \( f'(x) = 0 \):
\[ 3x^2 - 10x + 8 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} \]
\[ = \frac{10 \pm \sqrt{100 - 96}}{6} \]
\[ = \frac{10 \pm 2}{6} \]
This gives us:
\[ x = \frac{12}{6} = 2 \quad \text{and} \quad x = \frac{8}{6} = \frac{4}{3} \]

### Step 4: Test intervals
The critical points divide the number line into intervals. We will test the sign of \( f'(x) \) in the intervals \( (-\infty, \frac{4}{3}) \), \( \left(\frac{4}{3}, 2\right) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, \frac{4}{3}) \)**: Choose \( x = 0 \)
\[ f'(0) = 3(0)^2 - 10(0) + 8 = 8 > 0 \quad \text{(increasing)} \]

2. **Interval \( \left(\frac{4}{3}, 2\right) \)**: Choose \( x = 1.5 \)
\[ f'(1.5) = 3(1.5)^2 - 10(1.5) + 8 = 3(2.25) - 15 + 8 = 6.75 - 15 + 8 = -0.25 < 0 \quad \text{(decreasing)} \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ f'(3) = 3(3)^2 - 10(3) + 8 = 27 - 30 + 8 = 5 > 0 \quad \text{(increasing)} \]

### Conclusion
The function \( f(x) = (x-1)(x-2)^2 \) is:
- **Strictly increasing** on the intervals \( (-\infty, \frac{4}{3}) \) and \( (2, \infty) \).
- **Strictly decreasing** on the interval \( \left(\frac{4}{3}, 2\right) \).