Question #6d237

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Question #6d237

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Answer 1 · 2016-11-22T21:00:41

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

Explanation:

A function $f$ $f$ is said to be one to one if for any $x_{0}, x_{1}$ $x_0, x_1$ in the domain of $f$ $f$ , $f (x_{0}) = f (x_{1})$ $f(x_0) = f(x_1)$ implies $x_{0} = x_{1}$ $x_0 = x_1$ . In other words, there is only one element in the domain of $f$ $f$ that maps to any given element in the range of $f$ $f$ .

We will first show that $f (x) = 3 x - 1$ $f(x) = 3x-1$ has this property. Suppose $x_{0}$ $x_0$ and $x_{1}$ $x_1$ are real numbers such that $f (x_{0}) = f (x_{1})$ $f(x_0) = f(x_1)$ . Then

$3 x_{0} - 1 = 3 x_{1} - 1$ $3x_0-1 = 3x_1-1$

$\Rightarrow 3 x_{0} = 3 x_{1}$ $=> 3x_0 = 3x_1$

$\Rightarrow x_{0} = x_{1}$ $=> x_0 = x_1$

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

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

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

Set $x = f (y) = 3 y - 1$ $x = f(y) = 3y-1$

$\Rightarrow x + 1 = 3 y$ $=> x+1 = 3y$

$\Rightarrow y = \frac{x + 1}{3} = f^{- 1} (x)$ $=> y = (x+1)/3 = f^(-1)(x)$

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

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Question #6d237

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