98merrillv
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For which pairs of functions is f(g(x))=x?

A) f(x)=x^2 and g(x)=1/x

B) f(x)=2/x and g(x)=2/x

C) f(x)x-2/3 and g(x)=2-3x

D) f(x)1/2x-2 and g(x)=1/2x+2

Respuesta :

The correct answer is:

B) f(x)=2/x and g(x)=2/x

Explanation:

To perform this composition of functions, we take the value of g(x) for each choice and substitute it in place of x in f(x).  

For this one, we have f(x) = 2/x and g(x) = 2/x.  This means we replace x in f(x) with 2/x:

f(g(x)) = f(2/x) = 2/(2/x)

This can also be written as:

f(2/x) = 2÷(2/x) = (2/1)÷(2/x)

When dividing fractions, flip the second one and multiply:

f(2/x) = (2/1)×(x/2) = (2*x)/(1*2) = (2x)/2 = x

danielmaduroh

The right answer is  B) f(x)=2/x and g(x)=2/x


This is a problem of composition of functions.


[tex]The \ \mathbf{composition} \ of \ the \ function \ f \ with \ the \ function \ g \ is:\\ \\ (f \circ g)(x)=f(g(x)) \\ \\ The \ domain \ of \ (f \circ g) \ is \ the \ set \ of \ all \ x \ in \ the \ domain \ of \ g \\ such \ that \ g(x) \ is \ in \ the \ domain \ of \ f[/tex]


So, the only functions that satisfies this concept is B). To prove this, let's take the following steps:

[tex]f(x)=\frac{2}{x} \ and \ g(x)=\frac{2}{x} \\ \\ f(g(x))=\frac{2}{\frac{2}{x}}=\frac{2x}{2} \\ \\ \therefore \boxed{f(g(x))=x}[/tex]


Since this composition of functions gives us the identity function, then we can conclude that the function g is the inverse of f and vice versa.

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