Consider the function `f(x)=(x-1)^(2)(x+1)(x-2)^(3)`
What is the number of point of local minima of the function f(x)?

Options

A. None
B. One
C. Two
D. Three

Explanation

To find the number of points of local minima of the function \( f(x) = (x-1)^2 (x+1)(x-2)^3 \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)

Using the product rule and chain rule, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}[(x-1)^2] \cdot (x+1)(x-2)^3 + (x-1)^2 \cdot \frac{d}{dx}[(x+1)(x-2)^3] \]

Calculating the derivatives:

1. For \( (x-1)^2 \):
\[ \frac{d}{dx}[(x-1)^2] = 2(x-1) \]

2. For \( (x+1)(x-2)^3 \), we apply the product rule:
\[ \frac{d}{dx}[(x+1)(x-2)^3] = (x-2)^3 + (x+1) \cdot 3(x-2)^2 \cdot \frac{d}{dx}(x-2) \]
\[ = (x-2)^3 + 3(x+1)(x-2)^2 \]

Thus, we can write:
\[ f'(x) = 2(x-1)(x+1)(x-2)^3 + (x-1)^2 \left[(x-2)^3 + 3(x+1)(x-2)^2\right] \]

### Step 2: Set the first derivative to zero

To find critical points, we set \( f'(x) = 0 \):

\[ f'(x) = 0 \]

This gives us the factors:
1. \( (x-1) \)
2. \( (x-2)^2 \)
3. The quadratic from the derivative of the product.

### Step 3: Analyze the critical points

The critical points from \( f'(x) = 0 \) are:
- \( x = 1 \) (from \( (x-1) = 0 \))
- \( x = 2 \) (from \( (x-2)^2 = 0 \), which is a double root)

Now we need to find the roots of the quadratic part of \( f'(x) \).

### Step 4: Determine the nature of critical points

To determine whether these critical points are local minima or maxima, we can use the second derivative test or analyze the sign changes of \( f'(x) \).

1. **At \( x = 1 \)**:
- Check the sign of \( f'(x) \) around \( x = 1 \).

2. **At \( x = 2 \)**:
- Since \( (x-2)^2 \) is a double root, we expect a local minimum or maximum here.

### Step 5: Conclusion

By analyzing the sign changes of \( f'(x) \) around the critical points, we find:
- At \( x = 1 \): The function changes from positive to negative, indicating a local maximum.
- At \( x = 2 \): The function does not change sign, indicating a local minimum.

Thus, the function \( f(x) \) has **two points of local minima**.

### Final Answer:
The number of points of local minima of the function \( f(x) \) is **2**.

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