A function is matched below against an interval, where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

Options

A. `{:(,"Interval,"Function"),(,-oo,-4),(,x^(3)+6x^(2)+6):}`
B. `{:(,"Interval,"Function"),(,(-oo,1/3],3x^(2)-2x+1):}`
C. `{:(,"Interval,"Function"),(,[2,oo),(,2x^(3)-3x^(2)-12x+6):}`
D. `{:(,"Interval,"Function"),(,(-oo,oo),(,x^(3)-3x^(2)+3x+3):}`

Answer

To solve the problem of determining which function-interval pair is incorrectly matched regarding whether the function is increasing, we will analyze each given function and its derivative step by step.

### Step 1: Analyze the first function
**Function:** \( y = x^3 + 6x^2 + 6 \)

**Derivative:**
\[ \frac{dy}{dx} = 3x^2 + 12x \]

**Condition for Increasing:**
\[ 3x^2 + 12x \geq 0 \]

**Factoring:**
\[ 3x(x + 4) \geq 0 \]

**Finding Critical Points:**
The critical points are \( x = 0 \) and \( x = -4 \).

**Testing Intervals:**
- For \( x < -4 \): Choose \( x = -5 \) → \( 3(-5)(-5 + 4) = 15 > 0 \) (increasing)
- For \( -4 < x < 0 \): Choose \( x = -2 \) → \( 3(-2)(-2 + 4) = -12 < 0 \) (decreasing)
- For \( x > 0 \): Choose \( x = 1 \) → \( 3(1)(1 + 4) = 15 > 0 \) (increasing)

**Conclusion for Function 1:**
The function is increasing on the intervals \( (-\infty, -4] \) and \( [0, \infty) \).

### Step 2: Analyze the second function
**Function:** \( y = 3x^2 - 2x + 1 \)

**Derivative:**
\[ \frac{dy}{dx} = 6x - 2 \]

**Condition for Increasing:**
\[ 6x - 2 \geq 0 \implies x \geq \frac{1}{3} \]

**Conclusion for Function 2:**
The function is increasing on the interval \( [\frac{1}{3}, \infty) \).

### Step 3: Analyze the third function
**Function:** \( y = 2x^3 - 3x^2 - 12x + 6 \)

**Derivative:**
\[ \frac{dy}{dx} = 6x^2 - 6x - 12 \]

**Condition for Increasing:**
\[ 6(x^2 - x - 2) \geq 0 \implies x^2 - x - 2 \geq 0 \]

**Factoring:**
\[ (x - 2)(x + 1) \geq 0 \]

**Finding Critical Points:**
The critical points are \( x = 2 \) and \( x = -1 \).

**Testing Intervals:**
- For \( x < -1 \): Choose \( x = -2 \) → \( (-2 - 2)(-2 + 1) = 4 > 0 \) (increasing)
- For \( -1 < x < 2 \): Choose \( x = 0 \) → \( (0 - 2)(0 + 1) = -2 < 0 \) (decreasing)
- For \( x > 2 \): Choose \( x = 3 \) → \( (3 - 2)(3 + 1) = 4 > 0 \) (increasing)

**Conclusion for Function 3:**
The function is increasing on the intervals \( (-\infty, -1] \) and \( [2, \infty) \).

### Summary of Results
1. **Function 1**: Increasing on \( (-\infty, -4] \) and \( [0, \infty) \)
2. **Function 2**: Increasing on \( [\frac{1}{3}, \infty) \) (but incorrectly matched with \( (-\infty, \frac{1}{3}) \))
3. **Function 3**: Increasing on \( (-\infty, -1] \) and \( [2, \infty) \)

### Incorrectly Matched Pair
The incorrectly matched pair is **Function 2** with the interval \( (-\infty, \frac{1}{3}) \).