Answer:
(a) we are given two functions f(x) and g(x) as:
and
![g(x)=(4x+1)/(x)](https://img.qammunity.org/2020/formulas/mathematics/high-school/lw5iezvittgk1vg5i627uhc5xoxl13e1jx.png)
Now we find out the composition maps fog and gof as:
![fog=f(g(x))\\\\=f((4x+1)/(x))\\\\=(1)/((4x+1)/(x)-4 )\\ \\=(x)/(4x+1-4x)\\ \\=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/ijyb79fwxg5g9i22kb3ju1ngqm4bfa9jbr.png)
Hence,
![fog=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/hzmy7ugojojc5dgwhx6xwwgmysmbf4np2t.png)
similarly we find gof.
![gof(x)=g(f(x))\\\\=g((1)/(x-4))\\\\=(4((1)/(x-4))+1 )/((1)/(x-4))\\ \\=((4+x-4)/(x-4) )/((1)/(x-4))\\ \\=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/qiucu2pwmihj1c1xy37ju8rzy6aiibv1u1.png)
hence,
![gof=x](https://img.qammunity.org/2020/formulas/mathematics/high-school/66bcwcsjchfnr3kx2v2tebldlat2g2rqz4.png)
as fog=gof=x this means that f and g are inverse of each other.
(b) the domain of f is given by: (-∞,4)∪(4,∞)
since, f is defined all over the real line except at 4 ; at 4 the function is not defined as the denominator is zero.
also g is defined everywhere except at 0; since at 0 denominator is 0.
hence, domain of g(x) is (-∞,0)∪(0,∞).
domain of the composition fog is the domain of the function g(x) ( as for defining fog we need to evaluate function g(x) first and then f(g(x)) ).
and domain of function gof is equal to domain of function f(x) ( as for defining gof we need to evaluate function f(x) first and then g(f(x)) ).
hence domain of fog=(-∞,0)∪(0,∞).
and domain of gof=(-∞,4)∪(4,∞)