Use composition to determine if [tex]$G(x)[tex]$[/tex] or [tex]$[/tex]H(x)[tex]$[/tex] is the inverse of [tex]$[/tex]F(x)[tex]$[/tex] for the limited domain [tex]$[/tex]x \geq -1$[/tex].

[tex]\[
\begin{array}{l}
F(x) = \sqrt{x + 1} \\
G(x) = x^2 - 1 \\
H(x) = x^2 + 1
\end{array}
\][/tex]

A. [tex]$G(x)[tex]$[/tex] is the inverse of [tex]$[/tex]F(x)$[/tex].

B. [tex]$H(x)[tex]$[/tex] is the inverse of [tex]$[/tex]F(x)$[/tex].

C. Neither function is the inverse of [tex]$F(x)$[/tex].

D. Both functions are inverses of [tex]$F(x)$[/tex].



Answer :

To determine if \( G(x) \) or \( H(x) \) is the inverse of \( F(x) \) for the limited domain \( x \geq -1 \), we need to evaluate the compositions \( F(G(x)) \) and \( F(H(x)) \) and check if either composition equals \( x \).

Given:
[tex]\[ F(x) = \sqrt{x + 1} \][/tex]
[tex]\[ G(x) = x^2 - 1 \][/tex]
[tex]\[ H(x) = x^2 + 1 \][/tex]

### Composition of \( F \) and \( G \):

Let us first compute \( F(G(x)) \):
[tex]\[ F(G(x)) = F(x^2 - 1) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} \][/tex]

Since we are given the domain \( x \geq -1 \), \( x \) can be positive or zero. Therefore:
[tex]\[ \sqrt{x^2} = |x| \][/tex]

But to maintain \( F(x) \) as an inverse function, we need \( \sqrt{x^2} \) to simplify exactly to \( x \) for all \( x \geq -1 \). Since \( \sqrt{x^2} \) is \( |x| \), it does not always simplify to \( x \) without further restricting the domain.

### Composition of \( F \) and \( H \):

Now, let us compute \( F(H(x)) \):
[tex]\[ F(H(x)) = F(x^2 + 1) = \sqrt{(x^2 + 1) + 1} = \sqrt{x^2 + 2} \][/tex]

Examining \( \sqrt{x^2 + 2} \) for any \( x \geq -1 \), it does not simplify to \( x \).

### Evaluating the Results:

For \( F(G(x)) \):
[tex]\[ F(G(x)) = |x| \][/tex]
And for \( F(H(x)) \):
[tex]\[ F(H(x)) = \sqrt{x^2 + 2} \][/tex]

Neither composition results in \( x \). Thus, neither \( G(x) \) nor \( H(x) \) is an inverse of \( F(x) \).

Therefore, the correct answer is:

C. Neither function is the inverse of [tex]\( F(x) \)[/tex].