Find missing entires $f \circ g(x)$ calculus composite functions

Question

I have the calculus complete course book, edition 8 by Adams something. In chapter P5 exercise 15 I get a table with functions then I am supposed to find a missing one.

Question: If $f(x)=\frac{x+1}{x}$ and $f\circ g(x)=x$, then find $g(x)$.

I am supposed to find $g(x)$ but I do not know how. According to the solutions manual $g(x)=\frac{1}{x−1}$ but I do not know how to get that number.

Thank you for your help :)

Answer 1

Let $$f : x \mapsto \frac{x+1}{x} = 1+\frac{1}{x}$$

Then, for a given $x \in \mathbb{R}$,

$$ f(g(x))=x \Rightarrow 1+\frac{1}{g(x)}=x \text{ and } g(x)\ne0$$

$$\Rightarrow \frac{1}{g(x)}=x-1 \text{ and } g(x)\ne0$$

$$\Rightarrow g(x)=\frac{1}{x-1} \text{ and } g(x)\ne0 \text{ and } x\ne 1$$

Now, you must verify that $ \forall x \ne 1, (f\circ g)(x)=x$ and $g(x)\ne0$.

Answer 2

Hint: $$x=(f\circ g)(x)=f(g(x))=\frac{g(x)+1}{g(x)}\tag1$$

On base of $(1)$ find an expression of $g(x)$ in $x$.

Answer 3

$f(g(x))=x$

$\dfrac{g(x)+1}{g(x)}=x$

$xg(x)=g(x)+1$

$xg(x)-g(x)=1$

$g(x).(x-1)=1$

$g(x)=\dfrac{1}{x-1}$

Done!

Answer 4

Note that $f(x)=\dfrac{x+1}{x}$ is invertible and its inverse is found solving the equation wrt $x$ so that

$x f(x)=x+1\to x f(x)-x=1\to x(f(x)-1)=1\to x=\dfrac{1}{f(x)-1}$

thus $f^{-1}(x)=\dfrac{1}{x-1}$

Now back to the problem $(f\circ g)(x)=x$ apply $f^{-1}$ to both sides

$f^{-1}\circ (f\circ g)(x)=f^{-1}(x)$ associativity of function composition

$(f^{-1}\circ f)\circ g(x)=\dfrac{1}{x-1};\;(f^{-1}\circ f)=I$ identity

$g(x)=\dfrac{1}{x-1}$