To prove f(x)f(x) and f^-1(x)f−1(x) we need to show
(fcirc f^-1)(x)=x(f∘f−1)(x)=x and (f^-1circf)(x)=x(f−1∘f)(x)=x
f(x)=(2+x)/xf(x)=2+xx and f^-1(x) = 2/(x-1)f−1(x)=2x−1
(fcircf^-1)(x)(f∘f−1)(x)
= (2+2/(x-1))/(2/(x-1))=2+2x−12x−1
=((2(x-1))/(x-1)+2/(x-1))/(2/(x-1))=2(x−1)x−1+2x−12x−1
=(2(x-1)+2)/(x-1) xx (x-1)/2=2(x−1)+2x−1×x−12
=(2x-2+2)/(x-1)xx(x-1)/2=2x−2+2x−1×x−12
=(2x)/(x-1)xx(x-1)/2=2xx−1×x−12
=(cancel(2)x)/cancel(x-1)xxcancel(x-1)/cancel(2)
=x
We got that (fcirc f^-1)(x)=x
Similarly, we can prove (f^-1circf)(x)=x
(f^-1circf)(x)
=2/((2+x)/x - 1)
=2/((2+x)/x- x/x)
=2/((2+x-x)/x)
=2/((2+cancel(x)-cancel(x))/x)
=2/(2/x)
=2/1 xx x/2
=cancel(2)/1xxx/cancel(2)
=x
Therefore, (f^-1circf)(x)=x
Thus were are able to show that the functions are inverses of each other.
