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.

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(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:

$S = [-2,2]$

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(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:

$[1, \infty)$

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(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.

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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 $g(x)$.

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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!