Question

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 1

Answer:

Any value of x

Step-by-step explanation:

Given

$f(x) = \sqrt{x^2 - 1}$

$g(x) = \sqrt{x^2 + 1}$

Required

What value of x is  $f(g(x)) = g(f(x))$

Solving for f(g(x))

$f(x) = \sqrt{x^2 - 1}$

$f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}$

Solve the inner square

$f(g(x)) = \sqrt{(x^2 + 1 - 1}$

$f(g(x)) = \sqrt{x^2 } }$

$f(g(x)) = x$

Solving g(f(x))

$g(x) = \sqrt{x^2 + 1}$

$g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}$

$g(f(x)) = \sqrt{x^2 - 1 + 1}$

$g(f(x)) = \sqrt{x^2 }$

$g(f(x)) = x$

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

$f(g(x)) = g(f(x))$

$x = x$

This implies that $f(g(x)) = g(f(x))$ at any value of x