Let's Break it Down
A relation is any set of inputs and outputs. Some relations consist of sets of ordered pairs, and some relations can be expressed as equations. Some relations have one output for each input, and those relations are considered functions. Another name for inputs is domain, and another name for outputs is range.
When you create an equation, you apply mathematical operations to numbers and variables. For example, to build the equation \( y = 3x + 2 \), you start with x, multiply it by 3, and then add 2.
What if you had the equation \( y = 3x + 2 \), and you wanted to undo the operations you applied to x? Using algebraic properties, you would subtract 2, then divide by 3.
Consider the function below.
Data Set 1
| x | y |
|---|---|
| 0 | 1 |
| 1 | 3 |
| 2 | 5 |
| 3 | 7 |
Data Set 2
| x | y |
|---|---|
| 1 | 0 |
| 3 | 1 |
| 5 | 2 |
| 7 | 3 |
The domain and range of Data Set 1 are reversed in Data Set 2.
When two relations or functions are inverses of each other, the domain of the first is the range of the second, and the range of the first is the domain of the second. The notation for an inverse is a superscript -1, as \( f^{- 1}(x) \). Don't confuse the superscript -1 for a power—it is just the notation for the inverse.
Question
The two functions \( f\left( x \right) = 3x + 2 \) and \( f^{- 1}\left( x \right) = \frac{x - 2}{3} \) are inverses of each other. Find and completely simplify the composition of functions \( f(f^{- 1}(x)) \).
HINT: To find the composition of functions \( f(f^{- 1}(x)) \), evaluate \( f\left( \frac{x - 2}{3} \right) \).
\( f\left( f^{1}\left( x \right) \right) = 3\left( \frac{x - 2}{3} \right) + 2 \)
\( f\left( f^{1}\left( x \right) \right) = \require{enclose}\enclose{horizontalstrike}{3}\left( \frac{x - 2}{\require{enclose}\enclose{horizontalstrike}{3}} \right) + 2 \)
\( f\left( f^{1}\left( x \right) \right) = x - 2 + 2 \)
\( f\left( f^{1}\left( x \right) \right) = x \)
