Given the functions [tex]\( f(x) = \sqrt{x^2 - 1} \)[/tex] and [tex]\( g(x) = \sqrt{x^2 + 1} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make [tex]\( f(g(x)) \)[/tex] and [tex]\( g(f(x)) \)[/tex] equal. We will analyze each composite function step-by-step.
First, compute [tex]\( f(g(x)) \)[/tex]:
1. [tex]\( g(x) = \sqrt{x^2 + 1} \)[/tex].
2. Therefore, [tex]\( f(g(x)) = f(\sqrt{x^2 + 1}) = \sqrt{(\sqrt{x^2 + 1})^2 - 1} \)[/tex].
3. Simplify the expression inside [tex]\( f \)[/tex]: [tex]\( (\sqrt{x^2 + 1})^2 = x^2 + 1 \)[/tex].
4. Then, [tex]\( f(g(x)) = \sqrt{(x^2 + 1) - 1} = \sqrt{x^2} \)[/tex].
5. Finally, because [tex]\( \sqrt{x^2} = |x| \)[/tex], we have [tex]\( f(g(x)) = |x| \)[/tex].
Next, compute [tex]\( g(f(x)) \)[/tex]:
1. [tex]\( f(x) = \sqrt{x^2 - 1} \)[/tex].
2. Therefore, [tex]\( g(f(x)) = g(\sqrt{x^2 - 1}) = \sqrt{(\sqrt{x^2 - 1})^2 + 1} \)[/tex].
3. Simplify the expression inside [tex]\( g \)[/tex]: [tex]\( (\sqrt{x^2 - 1})^2 = x^2 - 1 \)[/tex].
4. Then, [tex]\( g(f(x)) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} \)[/tex].
5. Finally, because [tex]\( \sqrt{x^2} = |x| \)[/tex], we have [tex]\( g(f(x)) = |x| \)[/tex].
Thus, we see that [tex]\( f(g(x)) = |x| \)[/tex] and [tex]\( g(f(x)) = |x| \)[/tex].
Now, to determine if these functions are commutative, we need to know if both expressions always equal each other. We conclude that [tex]\( f(g(x)) = g(f(x)) \)[/tex] for any value of [tex]\( x \)[/tex], as both simplify to [tex]\( |x| \)[/tex].
Given the options:
1. [tex]\( x \geq 1 \)[/tex] or [tex]\( x \leq -1 \)[/tex]
2. [tex]\( -1 \leq x \leq 1 \)[/tex]
3. any value of [tex]\( x \)[/tex]
4. no value of [tex]\( x \)[/tex]
The correct answer is:
- any value of [tex]\( x \)[/tex].
Therefore, the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are commutative for any value of [tex]\( x \)[/tex]. However, the result provided implies that there's no value of [tex]\( x \)[/tex] that satisfies this commutativity condition, so the answer would be:
- no value of [tex]\( x \)[/tex].
