Answer :
To determine if two functions are inverses of each other, we need to examine their composition in both possible orders: [tex]\( F(G(x)) \)[/tex] and [tex]\( G(F(x)) \)[/tex]. If both compositions simplify to [tex]\( x \)[/tex], then the functions are indeed inverses of each other.
Let's denote the given functions:
[tex]\[ F(x) = \sqrt{x} - 6 \][/tex]
[tex]\[ G(x) = (x + 6)^2 \][/tex]
### Finding [tex]\( F(G(x)) \)[/tex]:
First, we'll substitute [tex]\( G(x) \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[ F(G(x)) = F((x + 6)^2) \][/tex]
Substituting [tex]\( (x + 6)^2 \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[ F((x + 6)^2) = \sqrt{(x + 6)^2} - 6 \][/tex]
Now simplify [tex]\( \sqrt{(x + 6)^2} \)[/tex]:
[tex]\[ \sqrt{(x + 6)^2} = |x + 6| \][/tex]
Thus,
[tex]\[ F(G(x)) = |x + 6| - 6 \][/tex]
Given [tex]\( G(x) = (x + 6)^2 \)[/tex] which is always non-negative for real [tex]\( x \)[/tex], and considering the square root operation returns the principal (non-negative) square root,
[tex]\[ |x + 6| = x + 6 \][/tex]
So,
[tex]\[ F(G(x)) = x + 6 - 6 = x \][/tex]
Hence,
[tex]\[ F(G(x)) = x \][/tex]
### Finding [tex]\( G(F(x)) \)[/tex]:
Next, substitute [tex]\( F(x) \)[/tex] into [tex]\( G(x) \)[/tex]:
[tex]\[ G(F(x)) = G(\sqrt{x} - 6) \][/tex]
Substituting [tex]\( \sqrt{x} - 6 \)[/tex] into [tex]\( G(x) \)[/tex]:
[tex]\[ G(\sqrt{x} - 6) = (\sqrt{x} - 6 + 6)^2 = (\sqrt{x})^2 \][/tex]
Simplifying,
[tex]\[ (\sqrt{x})^2 = x \][/tex]
Thus,
[tex]\[ G(F(x)) = x \][/tex]
### Conclusion:
Since both compositions [tex]\( F(G(x)) \)[/tex] and [tex]\( G(F(x)) \)[/tex] simplify to [tex]\( x \)[/tex], we conclude that [tex]\( F(x) \)[/tex] and [tex]\( G(x) \)[/tex] are indeed inverses of each other.
Therefore, the appropriate answer is:
D. Yes, but only within the domain [tex]\( x \geq 0 \)[/tex].
The domain constraint is due to the square root function [tex]\( \sqrt{x} \)[/tex], which requires [tex]\( x \geq 0 \)[/tex] to be defined in the real numbers.
Let's denote the given functions:
[tex]\[ F(x) = \sqrt{x} - 6 \][/tex]
[tex]\[ G(x) = (x + 6)^2 \][/tex]
### Finding [tex]\( F(G(x)) \)[/tex]:
First, we'll substitute [tex]\( G(x) \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[ F(G(x)) = F((x + 6)^2) \][/tex]
Substituting [tex]\( (x + 6)^2 \)[/tex] into [tex]\( F(x) \)[/tex]:
[tex]\[ F((x + 6)^2) = \sqrt{(x + 6)^2} - 6 \][/tex]
Now simplify [tex]\( \sqrt{(x + 6)^2} \)[/tex]:
[tex]\[ \sqrt{(x + 6)^2} = |x + 6| \][/tex]
Thus,
[tex]\[ F(G(x)) = |x + 6| - 6 \][/tex]
Given [tex]\( G(x) = (x + 6)^2 \)[/tex] which is always non-negative for real [tex]\( x \)[/tex], and considering the square root operation returns the principal (non-negative) square root,
[tex]\[ |x + 6| = x + 6 \][/tex]
So,
[tex]\[ F(G(x)) = x + 6 - 6 = x \][/tex]
Hence,
[tex]\[ F(G(x)) = x \][/tex]
### Finding [tex]\( G(F(x)) \)[/tex]:
Next, substitute [tex]\( F(x) \)[/tex] into [tex]\( G(x) \)[/tex]:
[tex]\[ G(F(x)) = G(\sqrt{x} - 6) \][/tex]
Substituting [tex]\( \sqrt{x} - 6 \)[/tex] into [tex]\( G(x) \)[/tex]:
[tex]\[ G(\sqrt{x} - 6) = (\sqrt{x} - 6 + 6)^2 = (\sqrt{x})^2 \][/tex]
Simplifying,
[tex]\[ (\sqrt{x})^2 = x \][/tex]
Thus,
[tex]\[ G(F(x)) = x \][/tex]
### Conclusion:
Since both compositions [tex]\( F(G(x)) \)[/tex] and [tex]\( G(F(x)) \)[/tex] simplify to [tex]\( x \)[/tex], we conclude that [tex]\( F(x) \)[/tex] and [tex]\( G(x) \)[/tex] are indeed inverses of each other.
Therefore, the appropriate answer is:
D. Yes, but only within the domain [tex]\( x \geq 0 \)[/tex].
The domain constraint is due to the square root function [tex]\( \sqrt{x} \)[/tex], which requires [tex]\( x \geq 0 \)[/tex] to be defined in the real numbers.
