To find the domain and range of the function \( f(x) = \cos^{-1} \left( \log \left[ \lfloor \sqrt{x^3 + 1} \rfloor \right] \right) \), we will follow these steps:
### Step 1: Determine the expression inside the logarithm
The function involves \( \lfloor \sqrt{x^3 + 1} \rfloor \). First, we need to analyze the expression \( \sqrt{x^3 + 1} \).
### Step 2: Find the range of \( \sqrt{x^3 + 1} \)
- As \( x \) varies, \( x^3 + 1 \) is always positive for all \( x \).
- The minimum value occurs when \( x = 0 \):
\[
\sqrt{0^3 + 1} = \sqrt{1} = 1
\]
- As \( x \) increases, \( x^3 + 1 \) increases without bound, thus \( \sqrt{x^3 + 1} \) also increases without bound.
### Step 3: Determine the greatest integer function
- The greatest integer function \( \lfloor \sqrt{x^3 + 1} \rfloor \) will take integer values starting from 1 (when \( x = 0 \)) and can take any integer value as \( x \) increases.
- Therefore, \( \lfloor \sqrt{x^3 + 1} \rfloor \) can take values \( n \) where \( n \geq 1 \).
### Step 4: Find the range of the logarithm
- The logarithm function \( \log(n) \) is defined for \( n > 0 \).
- Since \( \lfloor \sqrt{x^3 + 1} \rfloor \geq 1 \), we have:
\[
\log(\lfloor \sqrt{x^3 + 1} \rfloor) \geq \log(1) = 0
\]
- As \( n \) increases, \( \log(n) \) also increases without bound. Thus:
\[
\log(\lfloor \sqrt{x^3 + 1} \rfloor) \text{ can take any value } [0, \infty)
\]
### Step 5: Determine the domain of the cosine inverse function
- The function \( \cos^{-1}(y) \) is defined for \( y \) in the range \( [-1, 1] \).
- Therefore, we need:
\[
0 \leq \log(\lfloor \sqrt{x^3 + 1} \rfloor) \leq 1
\]
- This implies:
\[
1 \leq \lfloor \sqrt{x^3 + 1} \rfloor \leq e
\]
- Since \( \lfloor \sqrt{x^3 + 1} \rfloor \) can take integer values, the possible values are \( 1, 2 \).
### Step 6: Find the corresponding \( x \) values
- For \( \lfloor \sqrt{x^3 + 1} \rfloor = 1 \):
\[
1 \leq \sqrt{x^3 + 1} < 2 \implies 1 \leq x^3 + 1 < 4 \implies 0 \leq x^3 < 3 \implies 0 \leq x < \sqrt[3]{3}
\]
- For \( \lfloor \sqrt{x^3 + 1} \rfloor = 2 \):
\[
2 \leq \sqrt{x^3 + 1} < 3 \implies 4 \leq x^3 + 1 < 9 \implies 3 \leq x^3 < 8 \implies \sqrt[3]{3} \leq x < 2
\]
### Step 7: Combine the intervals for the domain
- Therefore, the domain of \( f(x) \) is:
\[
[0, \sqrt[3]{3}) \cup [\sqrt[3]{3}, 2) = [0, 2)
\]
### Step 8: Determine the range of \( f(x) \)
- When \( \lfloor \sqrt{x^3 + 1} \rfloor = 1 \):
\[
f(x) = \cos^{-1}(0) = \frac{\pi}{2}
\]
- When \( \lfloor \sqrt{x^3 + 1} \rfloor = 2 \):
\[
f(x) = \cos^{-1}(\log(2))
\]
- Therefore, the range of \( f(x) \) is:
\[
\left[ \cos^{-1}(\log(2)), \frac{\pi}{2} \right]
\]
### Final Answer
- **Domain**: \( [0, 2) \)
- **Range**: \( \left[ \cos^{-1}(\log(2)), \frac{\pi}{2} \right] \)
