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A function $f$ is one-to-one if and only if every horizontal line intersects the graph of $f$ no more than once.
For each of the following functions, use the horizontal line test to determine whether it is one-to-one.
Is the function $f$ graphed in the following image one-to-one?
No.
We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of $f$ to elements in the range of $f.$ The inverse function maps each element from the range of $f$ back to its corresponding element from the domain of $f.$ Therefore, to find the inverse function of a one-to-one function $f,$ given any $y$ in the range of $f,$ we need to determine which $x$ in the domain of $f$ satisfies $f(x)=y.$ Since $f$ is one-to-one, there is exactly one such value $x.$ We can find that value $x$ by solving the equation $f(x)=y$ for $x.$ Doing so, we are able to write $x$ as a function of $y$ where the domain of this function is the range of $f$ and the range of this new function is the domain of $f.$ Consequently, this function is the inverse of $f,$ and we write $x={f}^{\mathrm{-1}}(y).$ Since we typically use the variable $x$ to denote the independent variable and $y$ to denote the dependent variable, we often interchange the roles of $x$ and $y,$ and write $y={f}^{\mathrm{-1}}(x).$ Representing the inverse function in this way is also helpful later when we graph a function $f$ and its inverse ${f}^{\mathrm{-1}}$ on the same axes.
Find the inverse for the function $f\left(x\right)=3x-4.$ State the domain and range of the inverse function. Verify that ${f}^{\mathrm{-1}}(f(x))=x.$
Follow the steps outlined in the strategy.
Step 1. If $y=3x-4,$ then $3x=y+4$ and $x=\frac{1}{3}y+\frac{4}{3}.$
Step 2. Rewrite as $y=\frac{1}{3}x+\frac{4}{3}$ and let $y={f}^{\mathrm{-1}}\left(x\right).$
Therefore, ${f}^{\mathrm{-1}}\left(x\right)=\frac{1}{3}x+\frac{4}{3}.$
Since the domain of $f$ is $(\text{\u2212}\infty ,\infty ),$ the range of ${f}^{\mathrm{-1}}$ is $(\text{\u2212}\infty ,\infty ).$ Since the range of $f$ is $(\text{\u2212}\infty ,\infty ),$ the domain of ${f}^{\mathrm{-1}}$ is $(\text{\u2212}\infty ,\infty ).$
You can verify that ${f}^{\mathrm{-1}}(f(x))=x$ by writing
Note that for ${f}^{\mathrm{-1}}(x)$ to be the inverse of $f(x),$ both ${f}^{\mathrm{-1}}(f(x))=x$ and $f({f}^{\mathrm{-1}}(x))=x$ for all x in the domain of the inside function.
Find the inverse of the function $f\left(x\right)=3x\text{/}(x-2).$ State the domain and range of the inverse function.
${f}^{\mathrm{-1}}\left(x\right)=\frac{2x}{x-3}.$ The domain of ${f}^{\mathrm{-1}}$ is $\left\{x\right|x\ne 3\}.$ The range of ${f}^{\mathrm{-1}}$ is $\left\{y\right|y\ne 2\}.$
Let’s consider the relationship between the graph of a function $f$ and the graph of its inverse. Consider the graph of $f$ shown in [link] and a point $\left(a,b\right)$ on the graph. Since $b=f(a),$ then ${f}^{\mathrm{-1}}\left(b\right)=a.$ Therefore, when we graph ${f}^{\mathrm{-1}},$ the point $(b,a)$ is on the graph. As a result, the graph of ${f}^{\mathrm{-1}}$ is a reflection of the graph of $f$ about the line $y=x.$
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