Let's walk through the problem step-by-step to find the domain of each function and each composite function.
1. Define the Functions:
[tex]\( f(x) = \sqrt{x + 1} \)[/tex] and [tex]\( g(x) = x^2 \)[/tex]
2. Determine the Domain of Each Function:
- Domain of [tex]\( f(x) \)[/tex]:
[tex]\( f(x) = \sqrt{x + 1} \)[/tex]
The square root function is defined for non-negative values inside the square root. Thus, we require:
[tex]\[
x + 1 \geq 0 \implies x \geq -1
\][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is [tex]\([-1, \infty)\)[/tex].
- Domain of [tex]\( g(x) \)[/tex]:
[tex]\( g(x) = x^2 \)[/tex]
The function [tex]\( x^2 \)[/tex] is defined for all real numbers. There are no restrictions.
Therefore, the domain of [tex]\( g(x) \)[/tex] is [tex]\((-\infty, \infty)\)[/tex].
3. Determine the Composite Functions and Their Domains:
- Composite Function [tex]\( (f \circ g)(x) = f(g(x)) \)[/tex]:
By definition, [tex]\( (f \circ g)(x) = f(g(x)) = f(x^2) \)[/tex].
First, determine the domain of [tex]\( g(x) \)[/tex]:
[tex]\[
g(x) = x^2
\][/tex]
The output of [tex]\( g(x) \)[/tex] for any real [tex]\( x \)[/tex] is a non-negative value [tex]\( x^2 \geq 0 \)[/tex]. The result will range from [tex]\([0, \infty)\)[/tex].
Now, check that this result ([tex]\([0, \infty)\)[/tex]) is within the domain of [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = \sqrt{x + 1}
\][/tex]
To satisfy the domain of [tex]\( f(x) \)[/tex], [tex]\( x \geq -1 \)[/tex]. However, [tex]\( x^2 \)[/tex] ranges from 0 and higher, which is always [tex]\(\geq -1\)[/tex].
Thus, the domain of [tex]\( f(g(x)) = f(x^2) \)[/tex] is all real numbers.
Therefore, the domain of [tex]\( f \circ g \)[/tex] is [tex]\((-\infty, \infty)\)[/tex].
- Composite Function [tex]\( (g \circ f)(x) = g(f(x)) \)[/tex]:
By definition, [tex]\( (g \circ f)(x) = g(f(x)) = g(\sqrt{x+1}) \)[/tex].
First, determine the domain of [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = \sqrt{x + 1}
\][/tex]
The domain of [tex]\( f(x) \)[/tex] is [tex]\([-1, \infty)\)[/tex].
Now, check that the result of [tex]\( f(x) \)[/tex] ([tex]\(\sqrt{x + 1}\)[/tex]) is within the domain of [tex]\( g(x) \)[/tex]:
[tex]\[
g(x) = x^2
\][/tex]
The output of [tex]\( \sqrt{x+1} \)[/tex] for [tex]\( x \geq -1 \)[/tex] is non-negative, which is always defined for [tex]\( x^2 \)[/tex].
This means the composite function [tex]\( (g \circ f)(x) = (\sqrt{x + 1})^2 \)[/tex] will also be defined where [tex]\( x \geq -1 \)[/tex].
Therefore, the domain of [tex]\( g \circ f \)[/tex] is [tex]\([-1, \infty)\)[/tex].
Summary:
[tex]\[
\begin{array}{l}
\text{domain of } f : \quad [-1, \infty) \\
\text{domain of } g : \quad (-\infty, \infty) \\
\text{domain of } f \circ g : \quad (-\infty, \infty) \\
\text{domain of } g \circ f : \quad [-1, \infty)
\end{array}
\][/tex]
The domain of [tex]\( g \circ f \)[/tex] is [tex]\([-1, \infty)\)[/tex].
