Algebra help?

Question

Consider the following functions 

f(x) = x/ x+ 6 and g(x)= 6/x 

Find f(g(x)) and it's domain 

Find g(f(x)) and it's domain 

Find f(f(x)) and it's domain 

And 
Find g(g(x)) and it's domain

 

Answer 1

The functions are f(x) = x/(x - 6) and g(x) = 6/x.

(1).

f[g(x)] = f[6/x]                               [ Since g(x) = 6/x ]

          = (6/x)/[(6/x) - 6]                [ Since f(x) = x/(x - 6) ]

          = (6/x)/[(6 - 6x)/x]

          = 1/(1 - x).

The denominator not equals to zero, 1 - x ≠ 0 ⇒ x ≠ 1, because it is un defined.

The domain of f(g(x)) = {x ∈ R, x ≠ 1}, where R is any real number.

 

(2).

g[f(x)] = g[x/(x - 6)]                     [ Since f(x) = x/(x - 6) ]

          = 6 / [x/(x - 6)]                  [ Since g(x) = 6/x ]

          = 6(x - 6) / x

The denominator not equals to zero, x ≠ 0, because it is undefined.

The domain of g[f(x)] = {x ∈ R, x ≠ 0}, where R is any real number.

 

(3).

f[f(x)] = f[x/(x - 6)]                                       [ Since f(x) = x/(x - 6) ]

          = [x/(x - 6)] / { [x/(x - 6)] - 6}             [ Since f(x) = x/(x - 6) ]

          = x / [x - 6(x - 6)]

          = x / (- 5x + 36)

The denominator not equals to zero, - 5x + 36 ≠ 0 ⇒ x ≠ 36/5, because it is undefined.

The domain of f[f(x)] = {x ∈ R, x ≠ 36/5}, where R is any real number.

 

(4).

g[g(x)] = g[6/x]                                       [ Since g(x) = 6/x ]

          = 6 / [6/x]                                    [ Since g(x) = 6/x ]

          = x.

The above function represents the linear function, there is no restriction of the domain, because it has no negative square root and it has no denominator with variable x

The domain of g[g(x)] = {x ∈ R}, where R is any real number.