To find the domain of the function \( f(\log_2(x^2 + 3x - 2)) \) given that the domain of \( f(x) \) is [1, 3], we need to follow these steps:
### Step 1: Determine the conditions for the logarithm
The logarithmic function \( \log_2(x^2 + 3x - 2) \) is defined only when its argument is positive. Therefore, we need to ensure:
\[
x^2 + 3x - 2 > 0
\]
### Step 2: Solve the inequality
To solve the inequality \( x^2 + 3x - 2 > 0 \), we first find the roots of the equation \( x^2 + 3x - 2 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}
\]
Let \( r_1 = \frac{-3 - \sqrt{17}}{2} \) and \( r_2 = \frac{-3 + \sqrt{17}}{2} \).
### Step 3: Analyze the intervals
The roots divide the number line into intervals. We need to test each interval to determine where the inequality \( x^2 + 3x - 2 > 0 \) holds:
1. \( (-\infty, r_1) \)
2. \( (r_1, r_2) \)
3. \( (r_2, \infty) \)
Using test points from each interval, we find:
- For \( x < r_1 \), the expression is positive.
- For \( r_1 < x < r_2 \), the expression is negative.
- For \( x > r_2 \), the expression is positive.
Thus, the solution to \( x^2 + 3x - 2 > 0 \) is:
\[
x \in (-\infty, r_1) \cup (r_2, \infty)
\]
### Step 4: Apply the domain of \( f(x) \)
Next, since \( f(x) \) is defined for \( x \) in the interval [1, 3], we need to ensure that:
\[
1 \leq \log_2(x^2 + 3x - 2) \leq 3
\]
### Step 5: Convert the logarithmic inequalities
We convert the logarithmic inequalities:
1. From \( \log_2(x^2 + 3x - 2) \geq 1 \):
\[
x^2 + 3x - 2 \geq 2^1 \implies x^2 + 3x - 4 \geq 0
\]
Finding roots of \( x^2 + 3x - 4 = 0 \):
\[
x = \frac{-3 \pm \sqrt{3^2 + 16}}{2} = \frac{-3 \pm 5}{2} \implies x = 1 \text{ or } x = -4
\]
The solution is \( x \in (-\infty, -4) \cup [1, \infty) \).
2. From \( \log_2(x^2 + 3x - 2) \leq 3 \):
\[
x^2 + 3x - 2 \leq 2^3 \implies x^2 + 3x - 10 \leq 0
\]
Finding roots of \( x^2 + 3x - 10 = 0 \):
\[
x = \frac{-3 \pm \sqrt{3^2 + 40}}{2} = \frac{-3 \pm 7}{2} \implies x = 2 \text{ or } x = -5
\]
The solution is \( x \in [-5, 2] \).
### Step 6: Find the intersection of the intervals
Now, we need to find the intersection of the intervals:
- From \( x^2 + 3x - 4 \geq 0 \): \( (-\infty, -4) \cup [1, \infty) \)
- From \( x^2 + 3x - 10 \leq 0 \): \( [-5, 2] \)
The intersection is:
\[
[1, 2]
\]
### Final Answer
Thus, the domain of \( f(\log_2(x^2 + 3x - 2)) \) is:
\[
[1, 2]
\]
