Determine the inverse g(x) of the function f(x) = (1+(4/x)), stating its domain and range. Verify that f(g(x)) = g(f(x))=x and that g′(f(x)) = (1/(f′(x))?

Let f(x) = y

y = 1 + (4/x)

Now replace y with x and x with y and find equation for y

x = 1 + (4/y)

(x-1) = (4/y)

y = 4/(x-1)

This g(x), the inverse of f(x)

g(x)= 4/(x-1)

The domain will be all real numbers except when (x-1)=0 or x=1

So Domain = (-∞,1),(1,+∞)

And Range = (-∞,0),(0,+∞)

f(g(x)) = f(4/(x-1)) = 1 + 4/(4/(x-1)) = 1+(x-1) = x

g(f(x)) = g(1+(4/x)) = 4/((1+(4/x))-1) = 4/(4/x) = x

So we get f(g(x)) = g(f(x))

Notice the error in copying the next part of your question

It should be g'(f(x)) = 1/(f'(g(x)))

g'(f(x)) = d/dx (g(f(x))) = d/dx (x) = 1

f'(g(x)) = d/dx (f(g(x))) = d/dx (x) = 1

1/[f'(g(x))] = 1/1 = 1

g'(f(x)) = 1/f'(g(x)) ( Notice the error in copying your question)