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Sagot :
To determine if [tex]\( G(x) \)[/tex] or [tex]\( H(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex] for the domain [tex]\( x \geq -1 \)[/tex], we need to verify the inverse properties through composition of functions.
Let's start with the given functions:
[tex]\[ F(x) = \sqrt{x+1} \][/tex]
[tex]\[ G(x) = x^2 - 1 \][/tex]
[tex]\[ H(x) = x^2 + 1 \][/tex]
### Check if [tex]\( G(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex]:
To check if [tex]\( G(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex], we need to verify:
1. [tex]\( F(G(x)) = x \)[/tex]
2. [tex]\( G(F(x)) = x \)[/tex]
1. Checking [tex]\( F(G(x)) \)[/tex]:
[tex]\[ F(G(x)) = F(x^2 - 1) \][/tex]
[tex]\[ F(x^2 - 1) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = |x| \][/tex]
For [tex]\( F(G(x)) \)[/tex] to be [tex]\( x \)[/tex], we need [tex]\( |x| = x \)[/tex]. This is true only when [tex]\( x \geq 0 \)[/tex].
2. Checking [tex]\( G(F(x)) \)[/tex]:
[tex]\[ G(F(x)) = G(\sqrt{x+1}) \][/tex]
[tex]\[ G(\sqrt{x+1}) = (\sqrt{x+1})^2 - 1 = x + 1 - 1 = x \][/tex]
So, for [tex]\( x \geq -1 \)[/tex]:
- [tex]\( G(F(x)) = x \)[/tex]
- [tex]\( F(G(x)) = x \)[/tex] only if [tex]\( x \geq 0 \)[/tex], but we need it to hold for all [tex]\( x \geq -1 \)[/tex].
Because [tex]\( F(G(x)) \)[/tex] does not hold for all [tex]\( x \geq -1 \)[/tex], [tex]\( G(x) \)[/tex] is not the inverse of [tex]\( F(x) \)[/tex].
### Check if [tex]\( H(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex]:
To check if [tex]\( H(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex], we need to verify:
1. [tex]\( F(H(x)) = x \)[/tex]
2. [tex]\( H(F(x)) = x \)[/tex]
1. Checking [tex]\( F(H(x)) \)[/tex]:
[tex]\[ F(H(x)) = F(x^2 + 1) \][/tex]
[tex]\[ F(x^2 + 1) = \sqrt{(x^2 + 1) + 1} = \sqrt{x^2 + 2} \][/tex]
We need [tex]\( \sqrt{x^2 + 2} = x \)[/tex], which is never true for [tex]\( x \geq -1 \)[/tex].
2. Checking [tex]\( H(F(x)) \)[/tex]:
[tex]\[ H(F(x)) = H(\sqrt{x+1}) \][/tex]
[tex]\[ H(\sqrt{x+1}) = (\sqrt{x+1})^2 + 1 = x + 1 + 1 = x + 2 \][/tex]
We need [tex]\( x + 2 = x \)[/tex], which is never true for any [tex]\( x \)[/tex].
Since neither [tex]\( F(H(x)) = x \)[/tex] nor [tex]\( H(F(x)) = x \)[/tex] holds true, [tex]\( H(x) \)[/tex] is not the inverse of [tex]\( F(x) \)[/tex].
### Conclusion:
Since neither [tex]\( G(x) \)[/tex] nor [tex]\( H(x) \)[/tex] satisfies the conditions to be the inverse of [tex]\( F(x) \)[/tex], we can conclude:
B. Neither function is the inverse of [tex]\( F(x) \)[/tex].
Let's start with the given functions:
[tex]\[ F(x) = \sqrt{x+1} \][/tex]
[tex]\[ G(x) = x^2 - 1 \][/tex]
[tex]\[ H(x) = x^2 + 1 \][/tex]
### Check if [tex]\( G(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex]:
To check if [tex]\( G(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex], we need to verify:
1. [tex]\( F(G(x)) = x \)[/tex]
2. [tex]\( G(F(x)) = x \)[/tex]
1. Checking [tex]\( F(G(x)) \)[/tex]:
[tex]\[ F(G(x)) = F(x^2 - 1) \][/tex]
[tex]\[ F(x^2 - 1) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = |x| \][/tex]
For [tex]\( F(G(x)) \)[/tex] to be [tex]\( x \)[/tex], we need [tex]\( |x| = x \)[/tex]. This is true only when [tex]\( x \geq 0 \)[/tex].
2. Checking [tex]\( G(F(x)) \)[/tex]:
[tex]\[ G(F(x)) = G(\sqrt{x+1}) \][/tex]
[tex]\[ G(\sqrt{x+1}) = (\sqrt{x+1})^2 - 1 = x + 1 - 1 = x \][/tex]
So, for [tex]\( x \geq -1 \)[/tex]:
- [tex]\( G(F(x)) = x \)[/tex]
- [tex]\( F(G(x)) = x \)[/tex] only if [tex]\( x \geq 0 \)[/tex], but we need it to hold for all [tex]\( x \geq -1 \)[/tex].
Because [tex]\( F(G(x)) \)[/tex] does not hold for all [tex]\( x \geq -1 \)[/tex], [tex]\( G(x) \)[/tex] is not the inverse of [tex]\( F(x) \)[/tex].
### Check if [tex]\( H(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex]:
To check if [tex]\( H(x) \)[/tex] is the inverse of [tex]\( F(x) \)[/tex], we need to verify:
1. [tex]\( F(H(x)) = x \)[/tex]
2. [tex]\( H(F(x)) = x \)[/tex]
1. Checking [tex]\( F(H(x)) \)[/tex]:
[tex]\[ F(H(x)) = F(x^2 + 1) \][/tex]
[tex]\[ F(x^2 + 1) = \sqrt{(x^2 + 1) + 1} = \sqrt{x^2 + 2} \][/tex]
We need [tex]\( \sqrt{x^2 + 2} = x \)[/tex], which is never true for [tex]\( x \geq -1 \)[/tex].
2. Checking [tex]\( H(F(x)) \)[/tex]:
[tex]\[ H(F(x)) = H(\sqrt{x+1}) \][/tex]
[tex]\[ H(\sqrt{x+1}) = (\sqrt{x+1})^2 + 1 = x + 1 + 1 = x + 2 \][/tex]
We need [tex]\( x + 2 = x \)[/tex], which is never true for any [tex]\( x \)[/tex].
Since neither [tex]\( F(H(x)) = x \)[/tex] nor [tex]\( H(F(x)) = x \)[/tex] holds true, [tex]\( H(x) \)[/tex] is not the inverse of [tex]\( F(x) \)[/tex].
### Conclusion:
Since neither [tex]\( G(x) \)[/tex] nor [tex]\( H(x) \)[/tex] satisfies the conditions to be the inverse of [tex]\( F(x) \)[/tex], we can conclude:
B. Neither function is the inverse of [tex]\( F(x) \)[/tex].
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