Let `f(x)=(x-1)/(x^(2))` then which of the following is correct?

Options

A. f(x) has minima but no maxima
B. f(x) is inceases in the interval (0, 2) and decreases in the
C. f(x) is concave down in `(-oo, 0)uu(0, 3)`
D. x = 3 is the point of inflection

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x - 1}{x^2} \) in terms of its increasing/decreasing behavior, concavity, and points of inflection.

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

To determine where the function is increasing or decreasing, we first find the derivative:

\[ f'(x) = \frac{d}{dx} \left( \frac{x - 1}{x^2} \right) \]

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:

\[ f'(x) = \frac{(1)(x^2) - (x - 1)(2x)}{(x^2)^2} \]

Expanding 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, we have:

\[ f'(x) = \frac{2x - x^2}{x^4} = \frac{-x^2 + 2x}{x^4} \]

### Step 2: Set the first derivative to zero to find critical points

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

\[ -x^2 + 2x = 0 \]

Factoring out \( x \):

\[ x(-x + 2) = 0 \]

This gives us:

\[ x = 0 \quad \text{or} \quad x = 2 \]

### Step 3: Determine the intervals of increase and decrease

We analyze the sign of \( f'(x) \) in the intervals determined by the critical points \( x = 0 \) and \( x = 2 \):

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \):
\[ f'(-1) = \frac{-(-1)^2 + 2(-1)}{(-1)^4} = \frac{-1 - 2}{1} = -3 \quad (\text{decreasing}) \]

2. **Interval \( (0, 2) \)**: Choose \( x = 1 \):
\[ f'(1) = \frac{-1^2 + 2(1)}{1^4} = \frac{-1 + 2}{1} = 1 \quad (\text{increasing}) \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ f'(3) = \frac{-3^2 + 2(3)}{3^4} = \frac{-9 + 6}{81} = \frac{-3}{81} < 0 \quad (\text{decreasing}) \]

### Step 4: Conclusion on increasing/decreasing behavior

- \( f(x) \) is **decreasing** on \( (-\infty, 0) \) and \( (2, \infty) \).
- \( f(x) \) is **increasing** on \( (0, 2) \).

### Step 5: Find the second derivative \( f''(x) \)

To check for concavity and points of inflection, we differentiate \( f'(x) \):

\[ f'(x) = \frac{-x^2 + 2x}{x^4} \]

Using the quotient rule again:

\[ f''(x) = \frac{(-2x + 2)(x^4) - (-x^2 + 2x)(4x^3)}{(x^4)^2} \]

Calculating the numerator:

1. First term: \( (-2x + 2)x^4 \)
2. Second term: \( -(-x^2 + 2x)(4x^3) = (x^2 - 2x)(4x^3) = 4x^5 - 8x^4 \)

Combining these gives:

\[ f''(x) = \frac{-2x^5 + 2x^4 - 4x^5 + 8x^4}{x^8} = \frac{-6x^5 + 10x^4}{x^8} \]

### Step 6: Set the second derivative to zero for points of inflection

Setting \( f''(x) = 0 \):

\[ -6x^5 + 10x^4 = 0 \]

Factoring out \( 2x^4 \):

\[ 2x^4(-3x + 5) = 0 \]

This gives:

\[ x = 0 \quad \text{or} \quad x = \frac{5}{3} \]

Since \( x = 0 \) is not in the domain of \( f(x) \), we consider \( x = \frac{5}{3} \).

### Step 7: Determine concavity

To find where \( f''(x) \) changes sign, we can test intervals around \( x = \frac{5}{3} \):

1. **Interval \( (0, \frac{5}{3}) \)**: Choose \( x = 1 \):
\[ f''(1) = \frac{-6(1)^5 + 10(1)^4}{(1)^8} = 4 \quad (\text{concave up}) \]

2. **Interval \( (\frac{5}{3}, \infty) \)**: Choose \( x = 2 \):
\[ f''(2) = \frac{-6(2)^5 + 10(2)^4}{(2)^8} = \frac{-192 + 160}{256} = -\frac{32}{256} < 0 \quad (\text{concave down}) \]

### Final Results

- The function has a **local maximum** at \( x = 2 \).
- The function is **concave down** for \( x > \frac{5}{3} \).
- The point of inflection is at \( x = \frac{5}{3} \).

### Summary of Results

1. **Increasing**: \( (0, 2) \)
2. **Decreasing**: \( (-\infty, 0) \) and \( (2, \infty) \)
3. **Local Maximum**: \( x = 2 \)
4. **Concave Down**: \( \left(\frac{5}{3}, \infty\right) \)
5. **Point of Inflection**: \( x = \frac{5}{3} \)