You should substitute y for `f(x)` and `g(x)` such that:

`y = 3x + 1 => y - 1 = 3x => x = y/3 - 1/3`

Hence, the equation of inverse to the function `f(x)` is `f^(-1)(x) = x/3 - 1/3` .

`y = x + 2 => x = y - 2`

Hence, the equation of inverse to the function `g(x)` is `g^(-1)(x) =x-2` .

You need to evaluate `(f/g)(x) = (f(x))/(g(x))` such that:

`(f/g)(x) = (3x + 1)/(x + 2)`

You need to evaluate `(f/g)(x)` at `x = 1` such that:

`(f/g)(1) = (3*1 + 1)/(1 + 2) => (f/g)(1) =4/3`

**Hence, evaluating the inverses to the function `f(x)` and
`g(x)` yields `f^(-1)(x) = x/3 - 1/3` and `g^(-1)(x) =
x-2` and evaluating `(f/g)(x)` and `(f/g)(1)` yields
`(f/g)(x) = (3x + 1)/(x + 2) ` and `(f/g)(1) = 4/3` .**

By definition if `f^(-1)` is inverse of `f` then `f(f^(-1)(x))=x.`

Inverse of `f:`

`3f^(-1)(x)+1=x`

Next we proceed as if we are solving a linear equation.

`3f^(-1)(x)=x-1`

`f^(-1)(x)=(x-1)/(3)`

Similary inverse of `g:`

`g^(-1)(x)+2=x`

`g^(-1)(x)=x-2`

By definition `(f/g)(x)=(f(x))/(g(x))` hence

`(f/g)(x)=(3x+1)/(x+2)`.

So `(f/g)(1)=(3cdot1+1)/(1+2)=4/3`.

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