Determine the intervals in which the function `f(x)=(x-1)(x+1)^(2)` is increasing or decreasing. Find also the points at which the tangents to the curve are parallel to x - axis.

Answer

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

### Step 1: Find the derivative of the function
To analyze the increasing and decreasing behavior of the function, we first need to find its derivative \( f'(x) \).

Using the product rule:
\[ f'(x) = (x-1)'(x+1)^2 + (x-1)(x+1)^2' \]
Calculating the derivatives:
- \( (x-1)' = 1 \)
- \( (x+1)^2' = 2(x+1) \)

Now substituting back:
\[ f'(x) = 1 \cdot (x+1)^2 + (x-1) \cdot 2(x+1) \]
Simplifying:
\[ f'(x) = (x+1)^2 + 2(x-1)(x+1) \]
Now, expanding \( 2(x-1)(x+1) \):
\[ 2(x^2 - 1) = 2x^2 - 2 \]
Thus,
\[ f'(x) = (x+1)^2 + 2x^2 - 2 \]
Now, expanding \( (x+1)^2 \):
\[ (x+1)^2 = x^2 + 2x + 1 \]
So,
\[ f'(x) = x^2 + 2x + 1 + 2x^2 - 2 = 3x^2 + 2x - 1 \]

### Step 2: Set the derivative equal to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 3x^2 + 2x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 2, c = -1 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ x = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \]
Calculating the two possible values:
1. \( x = \frac{2}{6} = \frac{1}{3} \)
2. \( x = \frac{-6}{6} = -1 \)

### Step 3: Determine the intervals of increase and decrease
We will now test the intervals defined by the critical points \( x = -1 \) and \( x = \frac{1}{3} \).

The intervals to test are:
1. \( (-\infty, -1) \)
2. \( (-1, \frac{1}{3}) \)
3. \( (\frac{1}{3}, \infty) \)

**Testing the intervals:**
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ f'(-2) = 3(-2)^2 + 2(-2) - 1 = 12 - 4 - 1 = 7 \quad (\text{positive}) \]
- For \( -1 < x < \frac{1}{3} \) (e.g., \( x = 0 \)):
\[ f'(0) = 3(0)^2 + 2(0) - 1 = -1 \quad (\text{negative}) \]
- For \( x > \frac{1}{3} \) (e.g., \( x = 1 \)):
\[ f'(1) = 3(1)^2 + 2(1) - 1 = 3 + 2 - 1 = 4 \quad (\text{positive}) \]

### Conclusion on intervals:
- The function is **increasing** on \( (-\infty, -1) \) and \( (\frac{1}{3}, \infty) \).
- The function is **decreasing** on \( (-1, \frac{1}{3}) \).

### Step 4: Find points where the tangent is parallel to the x-axis
The tangent to the curve is parallel to the x-axis where \( f'(x) = 0 \). From our previous calculations, we found:
- \( x = -1 \)
- \( x = \frac{1}{3} \)

Now we find the corresponding \( y \)-coordinates:
1. For \( x = -1 \):
\[ f(-1) = (-1-1)(-1+1)^2 = (-2)(0) = 0 \quad \Rightarrow \quad (-1, 0) \]
2. For \( x = \frac{1}{3} \):
\[ f\left(\frac{1}{3}\right) = \left(\frac{1}{3}-1\right)\left(\frac{1}{3}+1\right)^2 = \left(-\frac{2}{3}\right)\left(\frac{4}{3}\right)^2 = -\frac{2}{3} \cdot \frac{16}{9} = -\frac{32}{27} \quad \Rightarrow \quad \left(\frac{1}{3}, -\frac{32}{27}\right) \]

### Final Answer:
- The function \( f(x) \) is increasing on the intervals \( (-\infty, -1) \) and \( (\frac{1}{3}, \infty) \).
- The function is decreasing on the interval \( (-1, \frac{1}{3}) \).
- The points where the tangent is parallel to the x-axis are \( (-1, 0) \) and \( \left(\frac{1}{3}, -\frac{32}{27}\right) \).