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Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make f(g(x)) and g(f(x)) commutative? x greater-than-or-equal-to 1 or x less-than-or-equal-to negative 1 Negative 1 less-than-or-equal-to x less-than-or-equal-to 1 any value of x no value of x

Answer :

MrRoyal

Answer:

Any value of x

Step-by-step explanation:

Given

[tex]f(x) = \sqrt{x^2 - 1}[/tex]

[tex]g(x) = \sqrt{x^2 + 1}[/tex]

Required

What value of x is  [tex]f(g(x)) = g(f(x))[/tex]

Solving for f(g(x))

[tex]f(x) = \sqrt{x^2 - 1}[/tex]

[tex]f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}[/tex]

Solve the inner square

[tex]f(g(x)) = \sqrt{(x^2 + 1 - 1}[/tex]

[tex]f(g(x)) = \sqrt{x^2 } }[/tex]

[tex]f(g(x)) = x[/tex]

Solving g(f(x))

[tex]g(x) = \sqrt{x^2 + 1}[/tex]

[tex]g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}[/tex]

[tex]g(f(x)) = \sqrt{x^2 - 1 + 1}[/tex]

[tex]g(f(x)) = \sqrt{x^2 }[/tex]

[tex]g(f(x)) = x[/tex]

Equate f(g(x)) and g(f(x))

[tex]f(g(x)) = g(f(x))[/tex]

[tex]x = x[/tex]

This implies that [tex]f(g(x)) = g(f(x))[/tex] at any value of x

Answer:

C, any value of x

Step-by-step explanation:

EDGE

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