Question

Consider the following functions.

1
1 + x
Show that f and g are inverse functions algebraically.
f(g(x)) = f(
f(x) =
11
g(f(x)) = g([
11
x ≥ 0; g(x)=¹-x, 0 X
X

Answer 1

`\boxed{\begin{aligned}f(g(x))&=f\left((1-x)/(x)\right)\\&=x\end{aligned}}`
`\boxed{\begin{aligned}g(f(x))&=g\left((1)/(1+x)\right)\\&=x\end{aligned}}`

The graph of f is the reflection of the graph of g in the line y = x.

Explanation:

To show that two functions are inverses, we need to show that:

• f(g(x)) = x, for all x in the domain of g(x), and
• g(f(x)) = x, for all x in the domain of f(x).

Given functions:

`f(x)=(1)/(1+x), \quad x\geq 0`

`g(x)=(1-x)/(x), \quad 0 < x\leq 1`

Find f(g(x)) by substituting x = g(x) into f(x):

`\begin{aligned}f(g(x))&=f\left((1-x)/(x)\right)\\\\&=(1)/(1+\left((1-x)/(x)\right))\\\\&=(1)/((x)/(x)+\left((1-x)/(x)\right))\\\\&=(1)/((x+1-x)/(x))\\\\&=(1)/((1)/(x))\\\\&=x\end{aligned}`

Find g(f(x)) by substituting x = f(x) into g(x):

`\begin{aligned}g(f(x))&=g\left((1)/(1+x)\right)\\\\&=(1-\left((1)/(1+x)\right))/(\left((1)/(1+x)\right))\\\\&=((1+x)/(1+x)-\left((1)/(1+x)\right))/(\left((1)/(1+x)\right))\\\\&=((1+x-1)/(1+x))/(\left((1)/(1+x)\right))\\\\&=((x)/(1+x))/(\left((1)/(1+x)\right))\\\\&=(x)/(1)\\\\&=x\end{aligned}`

Therefore, as f(g(x)) = x and g(f(x)) = x, we have successfully demonstrated that f(x) and g(x) are inverses under the given domain restrictions.

If we graph the two functions given their domain restrictions (see attached), we can see that:

• The graph of f is the reflection of the graph of g in the line y = x.