To find the domain of the function \( y = \sec^{-1}(x^2 + 3x + 1) \), we need to determine the values of \( x \) for which the expression inside the secant inverse function is valid. The secant inverse function, \( \sec^{-1}(\theta) \), is defined for \( \theta \leq -1 \) or \( \theta \geq 1 \).
### Step 1: Set up the inequalities
We need to solve the following two inequalities:
1. \( x^2 + 3x + 1 \geq 1 \)
2. \( x^2 + 3x + 1 \leq -1 \)
### Step 2: Solve the first inequality
Starting with the first inequality:
\[
x^2 + 3x + 1 \geq 1
\]
Subtracting 1 from both sides gives:
\[
x^2 + 3x \geq 0
\]
Factoring out \( x \):
\[
x(x + 3) \geq 0
\]
To find the critical points, set the equation to zero:
\[
x(x + 3) = 0
\]
This gives us the critical points \( x = 0 \) and \( x = -3 \).
### Step 3: Test intervals for the first inequality
We will test the sign of \( x(x + 3) \) in the intervals determined by the critical points:
- Interval \( (-\infty, -3) \)
- Interval \( (-3, 0) \)
- Interval \( (0, \infty) \)
1. For \( x < -3 \) (e.g., \( x = -4 \)):
\[
(-4)(-4 + 3) = (-4)(-1) = 4 \quad (\text{positive})
\]
2. For \( -3 < x < 0 \) (e.g., \( x = -1 \)):
\[
(-1)(-1 + 3) = (-1)(2) = -2 \quad (\text{negative})
\]
3. For \( x > 0 \) (e.g., \( x = 1 \)):
\[
(1)(1 + 3) = (1)(4) = 4 \quad (\text{positive})
\]
Thus, the solution to the first inequality is:
\[
x \in (-\infty, -3] \cup [0, \infty)
\]
### Step 4: Solve the second inequality
Now, we solve the second inequality:
\[
x^2 + 3x + 1 \leq -1
\]
Adding 1 to both sides gives:
\[
x^2 + 3x + 2 \leq 0
\]
Factoring the quadratic:
\[
(x + 1)(x + 2) \leq 0
\]
Finding the critical points:
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
### Step 5: Test intervals for the second inequality
We will test the sign of \( (x + 1)(x + 2) \) in the intervals determined by the critical points:
- Interval \( (-\infty, -2) \)
- Interval \( (-2, -1) \)
- Interval \( (-1, \infty) \)
1. For \( x < -2 \) (e.g., \( x = -3 \)):
\[
(-3 + 1)(-3 + 2) = (-2)(-1) = 2 \quad (\text{positive})
\]
2. For \( -2 < x < -1 \) (e.g., \( x = -1.5 \)):
\[
(-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 \quad (\text{negative})
\]
3. For \( x > -1 \) (e.g., \( x = 0 \)):
\[
(0 + 1)(0 + 2) = (1)(2) = 2 \quad (\text{positive})
\]
Thus, the solution to the second inequality is:
\[
x \in [-2, -1]
\]
### Step 6: Combine the intervals
Now we combine the solutions from both inequalities:
1. From the first inequality: \( (-\infty, -3] \cup [0, \infty) \)
2. From the second inequality: \( [-2, -1] \)
The combined domain is:
\[
x \in (-\infty, -3] \cup [-2, -1] \cup [0, \infty)
\]
### Final Answer
The domain of the function \( y = \sec^{-1}(x^2 + 3x + 1) \) is:
\[
(-\infty, -3] \cup [-2, -1] \cup [0, \infty)
\]
