# Fog, Gof, Fof, and GoG f(x) = x + (1/x) I'm having a problem with how to do these functions, I can find the domain if I find the right equation. Please help! f(x) = x + (1/x) g(x) = ...

Fog, Gof, Fof, and GoG

f(x) = x + (1/x)

g(x) = x^2 + 14 x + 2

*f*

*g*

(

*f*

*g*)(

*x*) = domain

(b)

*g*

*f*

(

*g*

*f*)(

*x*) = domain

### 1 Answer | Add Yours

a) You need to compose the functions f and g such that:

`(fog)(x) = f(g(x)) `

Since `f(x) = x + (1/x), ` hence, reasoning by analogy, yields:

`f(g(x)) = g(x) + 1/(g(x))`

Substituting `x^2 + 14x + 2` for g(x) yields:

`f(g(x)) =(x^2 + 14x + 2) + 1/(x^2 + 14x + 2)`

`f(g(x)) = ((x^2 + 14x + 2)^2 + 1)/(x^2 + 14x + 2)`

**Since the denominator (`x^2 + 14x + 2` ) never cancels, hence, the domain of the function `(fog)(x) = f(g(x))` is the real set R.**

b) You need to compose the functions f and g such that:

`(gof)(x) = g(f(x))`

Since `f(x) =x^2 + 14x + 2` , hence, reasoning by analogy, yields:

`g(f(x)) = (f(x))^2 + 14f(x) + 2`

Substituting `x + 1/x` for `f(x)` yields:

`g(f(x)) = ((x+1)/x)^2 + 14((x+1)/x) + 2`

`g(f(x)) = (x^2 + 2x + 1 + 14x^2 + 14x + 2x^2)/x^2`

`g(f(x)) = (17x^2 + 16x + 1)/x^2`

**Since the denominator 1 cancels for 1, hence, the domain of the function `(gof)(x) = g(f(x))` is the real set R, excepting 0.**