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Which statement verifies that [tex][tex]$f(x)$[/tex][/tex] and [tex][tex]$g(x)$[/tex][/tex] are inverses of each other?

A. [tex][tex]$f(g(x))=x$[/tex][/tex]
B. [tex][tex]$f(g(x))=x$[/tex][/tex] and [tex][tex]$g(f(x))=-x$[/tex][/tex]
C. [tex][tex]$f(g(x))=\frac{1}{g(f(x))}$[/tex][/tex]
D. [tex][tex]$f(g(x))=x$[/tex][/tex] and [tex][tex]$g(f(x))=x$[/tex][/tex]


Sagot :

To verify that two functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses of each other, we must check two conditions:

1. [tex]\( f(g(x)) = x \)[/tex]
2. [tex]\( g(f(x)) = x \)[/tex]

This means applying [tex]\( f \)[/tex] to [tex]\( g(x) \)[/tex] and [tex]\( g \)[/tex] to [tex]\( f(x) \)[/tex] should both return the original input [tex]\( x \)[/tex].

Let's evaluate the statements provided:

1. [tex]\( f(g(x)) = x \)[/tex]: This condition alone is necessary but not sufficient on its own to prove that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses.
2. [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = -x \)[/tex]: This is incorrect because both [tex]\( f(g(x)) \)[/tex] and [tex]\( g(f(x)) \)[/tex] must return [tex]\( x \)[/tex], not [tex]\(-x \)[/tex].
3. [tex]\( f(g(x)) = \frac{1}{g(f(x))} \)[/tex]: This expression does not prove that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses, as it complicates the relationship unnecessarily and does not adhere to the correct definitions.
4. [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex]: This condition correctly states that applying one function after the other (in either order) returns the original input [tex]\( x \)[/tex]. This is the precise definition required to verify that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses of each other.

Therefore, the correct statement that verifies [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses of each other is:
[tex]\[ f(g(x)) = x \ \text{and} \ g(f(x)) = x \][/tex]

Thus, the correct answer is:
[tex]\[ f(g(x)) = x \ \text{and} \ g(f(x)) = x \][/tex]