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To determine whether the functions [tex]\( f(x) = \sqrt{2x + 2} \)[/tex] and [tex]\( g(x) = \frac{x^2 - 2}{2} \)[/tex] are inverses, we need to check if both conditions [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex] hold true. Lets' verify each one step-by-step.
1. Verify [tex]\( f(g(x)) = x \)[/tex]:
Substitute [tex]\( g(x) = \frac{x^2 - 2}{2} \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left( \frac{x^2 - 2}{2} \right) \][/tex]
Now, apply the function [tex]\( f \)[/tex] to [tex]\( \frac{x^2 - 2}{2} \)[/tex]:
[tex]\[ f\left( \frac{x^2 - 2}{2} \right) = \sqrt{2 \left( \frac{x^2 - 2}{2} \right) + 2} \][/tex]
Simplify the expression inside the square root:
[tex]\[ 2 \left( \frac{x^2 - 2}{2} \right) + 2 = x^2 - 2 + 2 = x^2 \][/tex]
Thus:
[tex]\[ f\left( \frac{x^2 - 2}{2} \right) = \sqrt{x^2} = |x| \][/tex]
Therefore:
[tex]\[ f(g(x)) = |x| \][/tex]
For [tex]\( f(g(x)) \)[/tex] to be equal to [tex]\( x \)[/tex], [tex]\( x \)[/tex] must be non-negative because [tex]\(|x|\)[/tex] is always non-negative. This condition may not be true for all [tex]\( x \)[/tex].
2. Verify [tex]\( g(f(x)) = x\)[/tex]:
Substitute [tex]\( f(x) = \sqrt{2x + 2} \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(\sqrt{2x + 2}) \][/tex]
Now, apply the function [tex]\( g \)[/tex] to [tex]\( \sqrt{2x + 2} \)[/tex]:
[tex]\[ g(\sqrt{2x + 2}) = \frac{(\sqrt{2x + 2})^2 - 2}{2} \][/tex]
Simplify the expression:
[tex]\[ (\sqrt{2x + 2})^2 = 2x + 2 \][/tex]
Thus:
[tex]\[ g(\sqrt{2x + 2}) = \frac{2x + 2 - 2}{2} = \frac{2x}{2} = x \][/tex]
Therefore:
[tex]\[ g(f(x)) = x \][/tex]
Since [tex]\( g(f(x)) = x \)[/tex] for all [tex]\( x \)[/tex], but [tex]\( f(g(x)) = |x| \)[/tex], not necessarily [tex]\( x \)[/tex], the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not true inverses of each other.
The correct statement is:
A. Functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses because [tex]\( f(g(x)) \neq g(f(x)) \)[/tex].
Thus, the correct answer is:
A. Functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses because [tex]\( f(g(x)) \neq g(f(x)) \)[/tex].
1. Verify [tex]\( f(g(x)) = x \)[/tex]:
Substitute [tex]\( g(x) = \frac{x^2 - 2}{2} \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left( \frac{x^2 - 2}{2} \right) \][/tex]
Now, apply the function [tex]\( f \)[/tex] to [tex]\( \frac{x^2 - 2}{2} \)[/tex]:
[tex]\[ f\left( \frac{x^2 - 2}{2} \right) = \sqrt{2 \left( \frac{x^2 - 2}{2} \right) + 2} \][/tex]
Simplify the expression inside the square root:
[tex]\[ 2 \left( \frac{x^2 - 2}{2} \right) + 2 = x^2 - 2 + 2 = x^2 \][/tex]
Thus:
[tex]\[ f\left( \frac{x^2 - 2}{2} \right) = \sqrt{x^2} = |x| \][/tex]
Therefore:
[tex]\[ f(g(x)) = |x| \][/tex]
For [tex]\( f(g(x)) \)[/tex] to be equal to [tex]\( x \)[/tex], [tex]\( x \)[/tex] must be non-negative because [tex]\(|x|\)[/tex] is always non-negative. This condition may not be true for all [tex]\( x \)[/tex].
2. Verify [tex]\( g(f(x)) = x\)[/tex]:
Substitute [tex]\( f(x) = \sqrt{2x + 2} \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(\sqrt{2x + 2}) \][/tex]
Now, apply the function [tex]\( g \)[/tex] to [tex]\( \sqrt{2x + 2} \)[/tex]:
[tex]\[ g(\sqrt{2x + 2}) = \frac{(\sqrt{2x + 2})^2 - 2}{2} \][/tex]
Simplify the expression:
[tex]\[ (\sqrt{2x + 2})^2 = 2x + 2 \][/tex]
Thus:
[tex]\[ g(\sqrt{2x + 2}) = \frac{2x + 2 - 2}{2} = \frac{2x}{2} = x \][/tex]
Therefore:
[tex]\[ g(f(x)) = x \][/tex]
Since [tex]\( g(f(x)) = x \)[/tex] for all [tex]\( x \)[/tex], but [tex]\( f(g(x)) = |x| \)[/tex], not necessarily [tex]\( x \)[/tex], the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not true inverses of each other.
The correct statement is:
A. Functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses because [tex]\( f(g(x)) \neq g(f(x)) \)[/tex].
Thus, the correct answer is:
A. Functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are not inverses because [tex]\( f(g(x)) \neq g(f(x)) \)[/tex].
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