The range of the piecewise function
`f(x) ={:{(3(x-1)^2-2,"if",x lt 6),(-2x+5,"if",x ge6):}`

Options

A. `(-oo,oo)`
B. `(-oo,-7)`
C. `[-2,oo)`
D. `(-oo,-7] uu [-2,oo)`

Explanation

To find the range of the piecewise function

\[ f(x) = \begin{cases} 3(x-1)^2 - 2 & \text{if } x < 6 \\ -2x + 5 & \text{if } x \geq 6 \end{cases} \]

we will analyze each piece separately.

### Step 1: Analyze the first piece \( f(x) = 3(x-1)^2 - 2 \) for \( x < 6 \)

1. The function \( f(x) = 3(x-1)^2 - 2 \) is a quadratic function that opens upwards (since the coefficient of \( (x-1)^2 \) is positive).
2. The vertex of this parabola occurs at \( x = 1 \) (the value that makes \( (x-1)^2 \) equal to zero).
3. At \( x = 1 \):
\[ f(1) = 3(1-1)^2 - 2 = -2 \]
4. Since the parabola opens upwards, the minimum value of the function is \( -2 \) at \( x = 1 \). As \( x \) approaches \( 6 \) from the left, \( f(x) \) approaches:
\[ f(6) = 3(6-1)^2 - 2 = 3(5)^2 - 2 = 75 - 2 = 73 \]
5. Therefore, for \( x < 6 \), the range of this piece is:
\[ [-2, 73) \]

### Step 2: Analyze the second piece \( f(x) = -2x + 5 \) for \( x \geq 6 \)

1. The function \( f(x) = -2x + 5 \) is a linear function with a negative slope.
2. At \( x = 6 \):
\[ f(6) = -2(6) + 5 = -12 + 5 = -7 \]
3. As \( x \) increases beyond \( 6 \), the value of \( f(x) \) decreases without bound. Thus, as \( x \) approaches infinity, \( f(x) \) approaches \( -\infty \).
4. Therefore, for \( x \geq 6 \), the range of this piece is:
\[ (-\infty, -7] \]

### Step 3: Combine the ranges from both pieces

1. The range from the first piece is \( [-2, 73) \).
2. The range from the second piece is \( (-\infty, -7] \).
3. Combining these two ranges, we have:
\[ (-\infty, -7] \cup [-2, 73) \]

### Final Result

The overall range of the piecewise function is:
\[ (-\infty, -7] \cup [-2, 73) \]