Question

trgsrg

Answer 1

g(x)=f^-1 (x)

Answer 2

so g(f(x))=x=f(g(x))

Answer 3

Since this function is a one to one function, its inverse is a true function. Thus \(f^{-1}(f(x))=x=f(f^{-1}(x))\)

Answer 4

It is one to one because it is an increasing function. That is:\[
x_1<x_2\implies f(x_1)<f(x_2)
\]For any set of \(x_1,x_2\) in the domain.

Answer 5

@mathhelpnow There is no simple formula, but there are two methods:

1. Prove that it is always increasing/decreasing or and so it is one to one.

2. Start with \(f^{-1}(f(x))\) and show that it equals \(x\) and then show \(f(f^{-1}(x))\) simplifies to \(x\).

Answer 6

Do you want to do method 1 or 2?

Answer 7

Okay for the first one we start with \[
x_1<x_2
\]Then multiply both sides by \(10\):\[
10x_1<10x_2
\]Then add \(2\) to both sides\[
10x_1+2<10x_2+2
\]In other words: \[
10x_1+2=f(x_1) < f(x_2)=10x_2+2
\]

This proves \[
x_1<x_2\implies f(x_1)<f(x_2)
\]

Answer 8

It's always increasing, making it one to one.

Answer 9

Now I can show you method 2.

Answer 10

\[
f(g(x)) = 10(g(x))+2 = 10\left(\frac{x-2}{10}\right)+2
\]

Answer 11

\[
=(x-2)+2=x
\]

Answer 12

\[
g(f(x)) = \frac{f(x)-2}{10}=\frac{(10x+2)-2}{10}=\frac{10x}{10}=x
\]

Answer 13

Both methods have their faults.

In both cases:
You have to know the answer before you prove anything. It is easy to prove something that is false by accidentally making an algebraic error.

Answer 14

Since this function was a line, it could tell it was one to one.