Please just (c)fog-¹​

f(x) = 1/(1 - x)

g(x) = (x - 1)/x

The inverse functions f ⁻¹(x) and g ⁻¹(x), if they exist, are such that

f( f ⁻¹(x) ) = x

g( g ⁻¹(x) ) = x

So we have

f( f ⁻¹(x) ) = 1/(1 - f ⁻¹(x) ) = x

Solve for f ⁻¹(x) :

1 = (1 - f ⁻¹(x) ) x

1 - f ⁻¹(x) = 1/x

f ⁻¹(x) = 1 - 1/x

f ⁻¹(x) = (x - 1)/x

and so f ⁻¹(x) = g(x).

Similarly, you can solve for g ⁻¹(x) :

g( g ⁻¹(x) ) = (g ⁻¹(x) - 1) / g ⁻¹(x) = x

1 - 1/g ⁻¹(x) = x

1/g ⁻¹(x) = 1 - x

g ⁻¹(x) = 1/(1 - x)

So we know f(x) and g(x) are inverses of one another,

f ⁻¹(x) = g(x)

g ⁻¹(x) = f(x)

Then

(a) (f o g)(x) = x

(b) (g o f ⁻¹)(x) = g(g(x)) = (g(x) - 1)/g(x) = 1 - 1/g(x) = 1 - x/(x - 1) = 1/(1 - x)

(c) (f o g ⁻¹)(x) = f(f(x)) = 1/(1 - f(x) ) = 1/(1 - 1/(1 - x)) = 1/(x/(x - 1)) = (x - 1)/x

no he doesnt because "-16' is negatiive

step-by-step explanation:

hello umm i don't understand your question can you explain it more pls

step-by-step explanation: