Let f : S → R, f(x) = √ 4 − x 2 , where S is the set of all real values of x for which f(x) is defined. Let g: R → R, where g(x) = x 2 + 1. a Find S . b Find the range of f and the range of g. c State whether or not f ◦ g and g ◦ f are defined and give a reason for each assertion

We are given two functions:

1. \( f: S \to \mathbb{R}, \quad f(x) = \sqrt{4 - x^2} \)  
2. \( g: \mathbb{R} \to \mathbb{R}, \quad g(x) = x^2 + 1 \)

We will answer each part step by step.

---

### **(a) Finding \( S \)**
The domain \( S \) of \( f(x) \) consists of all real values of \( x \) for which \( f(x) \) is defined. Since \( f(x) \) contains a square root, the expression inside must be non-negative:

\[
4 - x^2 \geq 0
\]

\[
4 \geq x^2
\]

\[
-2 \leq x \leq 2
\]

Thus, the domain of \( f \) is:

\[
S = [-2,2]
\]

---

### **(b) Finding the Range of \( f \) and \( g \)**

#### **Range of \( f(x) = \sqrt{4 - x^2} \)**

Since \( x \) lies in \( [-2,2] \), we analyze \( f(x) \):

- At \( x = 0 \): \( f(0) = \sqrt{4 - 0} = 2 \)
- At \( x = \pm2 \): \( f(\pm2) = \sqrt{4 - 4} = 0 \)
- Since \( f(x) \) is continuous and symmetric about \( x = 0 \), the values range from 0 to 2.

Thus, the range of \( f \) is:

\[
[0,2]
\]

#### **Range of \( g(x) = x^2 + 1 \)**

Since \( x^2 \geq 0 \) for all real \( x \), we have:

\[
x^2 + 1 \geq 1
\]

Since \( x^2 \) can take any value in \( [0, \infty) \), the minimum value of \( g(x) \) is \( 1 \), and there is no upper bound. Hence, the range of \( g \) is:

\[
[1, \infty)
\]

---

### **(c) Determining if \( f \circ g \) and \( g \circ f \) are defined**

#### **Checking \( f \circ g \)**
The composition \( (f \circ g)(x) \) means evaluating \( f(g(x)) \):

\[
f(g(x)) = f(x^2 + 1) = \sqrt{4 - (x^2 + 1)^2}
\]

For \( f(g(x)) \) to be defined, we need:

\[
4 - (x^2 + 1)^2 \geq 0
\]

\[
4 \geq (x^2 + 1)^2
\]

Since \( g(x) = x^2 + 1 \) and \( g(x) \geq 1 \), we see that \( g(x) \geq 1 \) always, meaning \( (x^2 + 1)^2 \geq 1^2 = 1 \), so \( 4 - (x^2 + 1)^2 \) becomes negative for most \( x \). Thus, \( f(g(x)) \) is not defined for most \( x \), meaning \( f \circ g \) is **not defined**.

#### **Checking \( g \circ f \)**
The composition \( (g \circ f)(x) \) means evaluating \( g(f(x)) \):

\[
g(f(x)) = g(\sqrt{4 - x^2}) = (\sqrt{4 - x^2})^2 + 1
\]

\[
= 4 - x^2 + 1
\]

\[
= 5 - x^2
\]

Since \( x \) is in \( S = [-2,2] \), the function \( g(f(x)) \) is defined for all \( x \in [-2,2] \), meaning \( g \circ f \) **is defined**.

---

### **Final Answers**
1. **Domain \( S \)**: \( [-2,2] \).
2. **Range of \( f \)**: \( [0,2] \).
3. **Range of \( g \)**: \( [1, \infty) \).
4. **\( f \circ g \) is not defined** because \( g(x) \) produces values outside the domain of \( f \).
5. **\( g \circ f \) is defined** because \( f(x) \) always produces valid inputs for \( g(x) \).

---

### **Follow-up Questions**
1. How do we find the domain of composite functions in general?
2. Why does \( f \circ g \) fail to be defined?
3. What is the maximum value of \( g(f(x)) \) in \( g \circ f \)?
4. Can we modify \( f \) or \( g \) so that both compositions are defined?
5. What does the graph of \( f(x) \) and \( g(x) \) look like?

### **Tip**
When dealing with square roots in functions, always check that the expression inside the root is non-negative to determine the domain properly!

This problem involves finding the domain and range of two functions f(x) = √(4 - x²) and g(x) = x² + 1. It also explores whether the compositions f ◦ g and g ◦ f are defined, with a detailed explanation of the reasoning behind each assertion. Suitable for students in Grades 10-12.

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