Find the range of `((x-1)(x+2))/(x+3)`.

Answer

To find the range of the function \( f(x) = \frac{(x-1)(x+2)}{(x+3)} \), we can follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( y = \frac{(x-1)(x+2)}{(x+3)} \).

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(x + 3) = (x - 1)(x + 2) \]

### Step 3: Expand both sides
Expanding both sides:
\[ yx + 3y = x^2 + 2x - x - 2 \]
This simplifies to:
\[ yx + 3y = x^2 + x - 2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 0 = x^2 + x - yx - 2 - 3y \]
This can be rewritten as:
\[ x^2 + (1 - y)x + (-2 - 3y) = 0 \]

### Step 5: Identify coefficients for the quadratic equation
In the quadratic equation \( ax^2 + bx + c = 0 \):
- \( a = 1 \)
- \( b = 1 - y \)
- \( c = -2 - 3y \)

### Step 6: Apply the discriminant condition
For \( x \) to be real, the discriminant must be greater than or equal to zero:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values:
\[ (1 - y)^2 - 4(1)(-2 - 3y) \geq 0 \]

### Step 7: Simplify the discriminant
Expanding the discriminant:
\[ (1 - y)^2 + 8 + 12y \geq 0 \]
This simplifies to:
\[ 1 - 2y + y^2 + 8 + 12y \geq 0 \]
Combining like terms:
\[ y^2 + 10y + 9 \geq 0 \]

### Step 8: Factor the quadratic inequality
Factoring gives:
\[ (y + 9)(y + 1) \geq 0 \]

### Step 9: Determine the critical points
The critical points are \( y = -9 \) and \( y = -1 \).

### Step 10: Analyze the intervals
We analyze the sign of the product \( (y + 9)(y + 1) \) over the intervals:
1. \( y < -9 \)
2. \( -9 \leq y < -1 \)
3. \( y \geq -1 \)

- For \( y < -9 \): Both factors are negative, so the product is positive.
- For \( -9 < y < -1 \): One factor is positive and the other is negative, so the product is negative.
- For \( y > -1 \): Both factors are positive, so the product is positive.

### Step 11: Write the solution for the range
The inequality \( (y + 9)(y + 1) \geq 0 \) holds true for:
- \( y \leq -9 \) or \( y \geq -1 \)

Thus, the range of the function is:
\[ (-\infty, -9] \cup [-1, \infty) \]

### Final Answer
The range of the function \( f(x) = \frac{(x-1)(x+2)}{(x+3)} \) is:
\[ (-\infty, -9] \cup [-1, \infty) \]