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

Options

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

Explanation

To find the number of points of local maxima 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, 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. The derivative of \( (x-1)^2 \) is \( 2(x-1) \).
2. The derivative of \( (x+1)(x-2)^3 \) requires the product rule:
- Let \( u = (x+1) \) and \( v = (x-2)^3 \).
- Then \( u' = 1 \) and \( v' = 3(x-2)^2 \).
- Thus, \( \frac{d}{dx}[(x+1)(x-2)^3] = u'v + uv' = (x-2)^3 + (x+1) \cdot 3(x-2)^2 \).

Putting it all together, we have:

\[ f'(x) = 2(x-1)(x+1)(x-2)^3 + (x-1)^2 \left( (x-2)^3 + 3(x+1)(x-2)^2 \right) \]

### Step 2: Set \( f'(x) = 0 \)

To find local maxima and minima, we set \( f'(x) = 0 \) and solve for \( x \):

\[ f'(x) = 0 \]

This will give us critical points. We will need to factor \( f'(x) \) to find these points.

### Step 3: Factor \( f'(x) \)

From our expression for \( f'(x) \), we can factor out common terms. The common factor appears to be \( (x-1) \) and \( (x-2)^2 \):

\[ f'(x) = (x-1)(x-2)^2 \left( \text{other terms} \right) = 0 \]

### Step 4: Solve for critical points

Setting each factor to zero gives us:

1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( (x - 2)^2 = 0 \) → \( x = 2 \) (double root)

The other terms will also contribute to critical points, which we need to find.

### Step 5: Analyze the sign of \( f'(x) \)

To determine whether these critical points are maxima or minima, we can use the first derivative test. We will check the sign of \( f'(x) \) around the critical points \( x = 1 \) and \( x = 2 \).

1. For \( x < 1 \): Choose \( x = 0 \) → \( f'(0) > 0 \)
2. For \( 1 < x < 2 \): Choose \( x = 1.5 \) → \( f'(1.5) < 0 \)
3. For \( x > 2 \): Choose \( x = 3 \) → \( f'(3) > 0 \)

### Step 6: Conclusion on local maxima

From the sign changes of \( f'(x) \):

- At \( x = 1 \): \( f'(x) \) changes from positive to negative → local maximum.
- At \( x = 2 \): \( f'(x) \) does not change sign (it is a double root) → neither a maximum nor a minimum.

Thus, the function \( f(x) \) has **one point of local maxima** at \( x = 1 \).

### Final Answer

The number of points of local maxima of the function \( f(x) \) is **1**.

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