the function `(|x-1|)/x^2` is monotonically decreasing at the point

Options

A. `x=3`
B. `x=1`
C. `x=2`
D. none of these

Answer

To determine the values of \( x \) for which the function \( f(x) = \frac{|x-1|}{x^2} \) is monotonically decreasing, we need to find the derivative of the function and analyze where it is negative.

### Step 1: Define the function
Given the function:
\[ f(x) = \frac{|x-1|}{x^2} \]

### Step 2: Analyze the absolute value
The absolute value function \( |x-1| \) can be expressed in two cases:
1. For \( x \geq 1 \): \( |x-1| = x-1 \)
2. For \( x < 1 \): \( |x-1| = -(x-1) = 1-x \)

Since we are interested in the monotonicity for \( x \geq 1 \) (as the context suggests), we will use the first case:
\[ f(x) = \frac{x-1}{x^2} \quad \text{for } x \geq 1 \]

### Step 3: Differentiate the function
Using the quotient rule, where \( u = x-1 \) and \( v = x^2 \):
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
Calculating \( u' \) and \( v' \):
- \( u' = 1 \)
- \( v' = 2x \)

Now substituting into the quotient rule:
\[ f'(x) = \frac{(1)(x^2) - (x-1)(2x)}{(x^2)^2} \]
Simplifying the numerator:
\[ f'(x) = \frac{x^2 - (2x^2 - 2x)}{x^4} = \frac{x^2 - 2x^2 + 2x}{x^4} = \frac{-x^2 + 2x}{x^4} \]
Thus:
\[ f'(x) = \frac{2x - x^2}{x^4} = \frac{x(2 - x)}{x^4} \]

### Step 4: Determine where the derivative is negative
For \( f(x) \) to be monotonically decreasing, we need:
\[ f'(x) < 0 \]
This occurs when:
\[ x(2 - x) < 0 \]
The critical points occur when \( x = 0 \) and \( x = 2 \).

### Step 5: Analyze the intervals
We analyze the sign of \( f'(x) \) in the intervals defined by the critical points:
1. \( x < 0 \): Not applicable since we are considering \( x \geq 1 \).
2. \( 0 < x < 2 \): \( f'(x) > 0 \) (not applicable).
3. \( x = 2 \): \( f'(2) = 0 \).
4. \( x > 2 \): \( f'(x) < 0 \).

### Conclusion
The function \( f(x) \) is monotonically decreasing for \( x > 2 \).

### Final Answer
The function \( f(x) = \frac{|x-1|}{x^2} \) is monotonically decreasing at the point \( x = 2 \) and for all \( x > 2 \).

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