The function `f(x)=[log (x-1)]^(2) (x-1)^(2)` has :

Options

A. Local extremum at x=1
B. Point of inflection at x=1
C. Local extremum at x=2
D. Point of inflection at x=2

Answer

To find the extremum of the function \( f(x) = [\log(x-1)]^2 (x-1)^2 \), we will follow these steps:

### Step 1: Differentiate the Function
We will use the product rule and chain rule to differentiate the function \( f(x) \).

Let:
- \( u = [\log(x-1)]^2 \)
- \( v = (x-1)^2 \)

Using the product rule:
\[ f'(x) = u'v + uv' \]

#### Differentiating \( u \):
Using the chain rule:
\[ u' = 2[\log(x-1)] \cdot \frac{1}{x-1} \]

#### Differentiating \( v \):
\[ v' = 2(x-1) \]

Putting it all together:
\[ f'(x) = \left(2[\log(x-1)] \cdot \frac{1}{x-1}\right)(x-1)^2 + [\log(x-1)]^2(2(x-1)) \]
This simplifies to:
\[ f'(x) = 2(x-1)[\log(x-1)] + 2(x-1)^2[\log(x-1)]^2 \]

### Step 2: Set the Derivative to Zero
To find the critical points, we set \( f'(x) = 0 \):
\[ 2(x-1)[\log(x-1)] + 2(x-1)^2[\log(x-1)]^2 = 0 \]
Factoring out \( 2(x-1) \):
\[ 2(x-1)\left[\log(x-1) + (x-1)[\log(x-1)]^2\right] = 0 \]

This gives us two cases:
1. \( 2(x-1) = 0 \) which gives \( x = 1 \)
2. \( \log(x-1) + (x-1)[\log(x-1)]^2 = 0 \)

### Step 3: Solve for Critical Points
For the second case, we can analyze:
\[ \log(x-1) + (x-1)[\log(x-1)]^2 = 0 \]
Let \( y = \log(x-1) \), then we rewrite the equation:
\[ y + (e^y - 1)y^2 = 0 \]
This equation can be solved numerically or graphically to find the roots.

### Step 4: Identify the Roots
From the analysis:
1. \( x = 1 \) is not a valid root since \( \log(0) \) is undefined.
2. For \( y = 0 \) (which corresponds to \( x = 2 \)), we find that \( \log(1) = 0 \).
3. For \( y = -1 \) (which corresponds to \( x = 1/e + 1 \)), we find that \( \log(1/e) = -1 \).

### Step 5: Determine the Nature of Critical Points
To determine whether these critical points are maxima or minima, we can use the second derivative test or analyze the sign of \( f'(x) \) around the critical points.

### Conclusion
The function \( f(x) \) has local extrema at:
- \( x = 2 \)
- \( x = 1/e + 1 \)

Thus, the function has two extremum points.