# Question #6d237

Nov 22, 2016

$f \left(x\right) = 3 x - 1$ is one to one, and has the inverse ${f}^{- 1} \left(x\right) = \frac{x + 1}{3}$

#### Explanation:

A function $f$ is said to be one to one if for any ${x}_{0} , {x}_{1}$ in the domain of $f$, $f \left({x}_{0}\right) = f \left({x}_{1}\right)$ implies ${x}_{0} = {x}_{1}$. In other words, there is only one element in the domain of $f$ that maps to any given element in the range of $f$.

We will first show that $f \left(x\right) = 3 x - 1$ has this property. Suppose ${x}_{0}$ and ${x}_{1}$ are real numbers such that $f \left({x}_{0}\right) = f \left({x}_{1}\right)$. Then

$3 {x}_{0} - 1 = 3 {x}_{1} - 1$

$\implies 3 {x}_{0} = 3 {x}_{1}$

$\implies {x}_{0} = {x}_{1}$

Thus $f \left(x\right) = 3 x - 1$ is one to one.

As $f \left(x\right)$ is one to one, it has an inverse function ${f}^{- 1} \left(x\right)$ where ${f}^{- 1} \left(f \left(x\right)\right) = f \left({f}^{- 1} \left(x\right)\right) = x$.

One way of finding ${f}^{- 1}$ is to set $y = f \left(x\right) = 3 x - 1$, change all the $x$'s to $y$'s and vice versa, giving $x = f \left(y\right) = 3 y - 1$, and then solve for $y$. This results in an equation of the form $y = {f}^{- 1} \left(x\right)$.

Set $x = f \left(y\right) = 3 y - 1$

$\implies x + 1 = 3 y$

$\implies y = \frac{x + 1}{3} = {f}^{- 1} \left(x\right)$

Notice that given ${f}^{- 1} \left(x\right) = \frac{x + 1}{3}$ we have ${f}^{- 1} \left(f \left(x\right)\right) = f \left({f}^{- 1} \left(x\right)\right) = x$, as desired.