# For the functions f(x)=1/(x)^2 g(x)=7x-6 h(x)=1-2x a) fog(x) c)hof(x) d)gofoh(x) State restrictions where possible i.e for rational function

*print*Print*list*Cite

### 1 Answer

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

`(fog)(x) = f(g(x)) = 1/(g^2(x))`

You need to substitute 7x - 6 for g(x) such that:

`(fog)(x) = f(g(x)) = 1/((7x - 6)^2)`

The denominator of the fraction needs to be different from zero, hence `x!=6/7` .

Thus, the domain of the function needs to exclude the value 6/7.

**Hence, evaluating `(fog)(x)` yields `(fog)(x) = 1/((7x - 6)^2).` **

c) You need to compose the functions h(x) and f(x) such that:

`(hof)(x) = h(f(x)) = 1 - 2f(x)`

You need to substitute `1/(x^2)` for f(x) such that:

`(hof)(x) = h(f(x)) = 1 - 2/(x^2)`

Notice that the domain of the function needs to exclude the value 0.

**Hence, evaluating `(hof)(x)` yields `(hof)(x) = 1 - 2/(x^2).` **

d) `(gofoh)(x) = g(f(h(x))) = g(f(1-2x)) = 7f(1-2x) - 6`

`(gofoh)(x) = 7/((1-2x)^2) - 6`

Notice that the domain of the function needs to exclude the value `1/2.`

**Hence, evaluating the `(gofoh)(x` ) yields`(gofoh)(x) = 7/((1-2x)^2) - 6` **