To solve the problem, we need to find the functions \( f + g \), \( f - g \), \( fg \), and \( f/g \) given the functions \( f(x) = \sqrt{1 - x^2} \) and \( g(x) = x^3 + 1 \).
### Step 1: Find \( f + g \)
The function \( f + g \) is defined as:
\[
f + g = f(x) + g(x) = \sqrt{1 - x^2} + (x^3 + 1)
\]
Thus,
\[
f + g = \sqrt{1 - x^2} + x^3 + 1
\]
**Domain:** The domain of \( f(x) \) is \( [-1, 1] \) and the domain of \( g(x) \) is \( \mathbb{R} \). The overall domain is the intersection of these two domains, which is \( [-1, 1] \).
**Range:** The range of \( f(x) \) is \( [0, 1] \) (since \( \sqrt{1 - x^2} \) varies from 0 to 1), and the range of \( g(x) \) is \( \mathbb{R} \). Therefore, the range of \( f + g \) is \( [1, 2] \) since at \( x = 0 \), \( f(0) + g(0) = 1 + 1 = 2 \) and at \( x = \pm 1 \), \( f(\pm 1) + g(\pm 1) = 0 + 2 = 2 \).
### Step 2: Find \( f - g \)
The function \( f - g \) is defined as:
\[
f - g = f(x) - g(x) = \sqrt{1 - x^2} - (x^3 + 1)
\]
Thus,
\[
f - g = \sqrt{1 - x^2} - x^3 - 1
\]
**Domain:** The domain remains \( [-1, 1] \).
**Range:** The minimum value occurs at \( x = 0 \) where \( f(0) - g(0) = 1 - 1 = 0 \). The maximum occurs at \( x = \pm 1 \) where \( f(\pm 1) - g(\pm 1) = 0 - 2 = -2 \). Therefore, the range of \( f - g \) is \( [-2, 0] \).
### Step 3: Find \( fg \)
The function \( fg \) is defined as:
\[
fg = f(x) \cdot g(x) = \sqrt{1 - x^2} \cdot (x^3 + 1)
\]
Thus,
\[
fg = \sqrt{1 - x^2} \cdot (x^3 + 1)
\]
**Domain:** The domain remains \( [-1, 1] \).
**Range:** The range can be determined by evaluating the function at critical points within the interval. The maximum occurs at \( x = 0 \) where \( fg(0) = 1 \cdot 1 = 1 \). The minimum occurs at \( x = \pm 1 \) where \( fg(\pm 1) = 0 \cdot 2 = 0 \). Therefore, the range of \( fg \) is \( [0, 1] \).
### Step 4: Find \( f/g \)
The function \( f/g \) is defined as:
\[
f/g = \frac{f(x)}{g(x)} = \frac{\sqrt{1 - x^2}}{x^3 + 1}
\]
Thus,
\[
f/g = \frac{\sqrt{1 - x^2}}{x^3 + 1}
\]
**Domain:** The domain is \( [-1, 1] \) since \( g(x) \) is never zero in this interval (as \( g(x) = x^3 + 1 \) is always positive for \( x \in [-1, 1] \)).
**Range:** The maximum occurs at \( x = 0 \) where \( f/g(0) = \frac{1}{1} = 1 \). The minimum occurs at \( x = \pm 1 \) where \( f/g(\pm 1) = 0 \). Therefore, the range of \( f/g \) is \( [0, 1] \).
### Summary of Results
- \( f + g = \sqrt{1 - x^2} + x^3 + 1 \) with domain \( [-1, 1] \) and range \( [1, 2] \).
- \( f - g = \sqrt{1 - x^2} - x^3 - 1 \) with domain \( [-1, 1] \) and range \( [-2, 0] \).
- \( fg = \sqrt{1 - x^2} \cdot (x^3 + 1) \) with domain \( [-1, 1] \) and range \( [0, 1] \).
- \( f/g = \frac{\sqrt{1 - x^2}}{x^3 + 1} \) with domain \( [-1, 1] \) and range \( [0, 1] \).
