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The inverse of a function can itself be a function. In this section, properties of inverses and *invertible functions* will be explored.

The inverse of a function reverses its $x$-$y$ coordinates. If, for a function $f,$ $x$ is an input and $y$ is its corresponding output, for the inverse, $f^{\text{-} 1}, y$ is the input and $x$ would be the corresponding output. $f(x)=y \quad \Leftrightarrow \quad f^{\text{-} 1}(y)=x$

Some function families are inverses of each other. This is because some functions undo each other. For example, $x^2$ and $\pm\sqrt{x}$ are inverses because radicals and exponents (with the same index) undo each other.
Depending on how a function is presented, finding its inverse can be done in different ways. When a function's rule is given, finding the inverse algebraically is advantageous. Consider the function $f(x)= \dfrac{2x-1}{3}.$
### 1

To begin, since $f(x)=y$ describes the input-output relationship of the function, replace $f(x)$ with $y$ in the function rule. $f(x)= \dfrac{2x-1}{3} \quad \Leftrightarrow \quad y= \dfrac{2x-1}{3}$

### 2

Because the inverse of a function reverses $x$ and $y,$ the variables can be switched. Notice that every other piece in the function rule remains the same. ${\color{#009600}{y}}= \dfrac{2 {\color{#0000FF}{x}}-1}{3} \quad \Rightarrow {\color{#0000FF}{x}}= \dfrac{2 {\color{#009600}{y}}-1}{3}$

### 3

Solve the resulting equation from Step 2 for $y.$ Here, this will involve using the inverse operations.
### 4

Replace $f(x)$ with $y$

Switch $x$ and $y$

Solve for $y$

$x=\dfrac{2y-1}{3}$

MultEqn$\text{LHS} \cdot 3=\text{RHS}\cdot 3$

$3x=2y-1$

AddEqn$\text{LHS}+1=\text{RHS}+1$

$3x+1=2y$

DivEqn$\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.$

$\dfrac{3x+1}{2}=y$

RearrangeEqnRearrange equation

$y=\dfrac{3x+1}{2}$

Replace $y$ with $f^{\text{-} 1}(x)$

Just as $f(x)=y$ shows the input-output relationship of $f,$ so does $f^{\text{-} 1}(x)=y.$ Thus, replacing $y$ with $f^{\text{-} 1}(x)$ gives the rule for the inverse of $f.$

$y=\dfrac{3x+1}{2} \quad \Leftrightarrow \quad f^{\text{-} 1}=\dfrac{3x+1}{2}$

Notice that in $f,$ the input is multiplied by $2,$ decreased by $1$ and divided by $3.$ From the rule of $f^{\text{-} 1},$ it can be seen that $x$ undergoes the inverse of these operation in the reverse order. Specifically, $x$ is multiplied by $3,$ increased by $1,$ and divided by $2.$

Some of the coordinates of the function $g$ are shown in the table. Find $g^{\text{-} 1},$ then graph $g$ and $g^{\text{-} 1}$ on the same coordinate plane.

$x$ | $g(x)$ |
---|---|

$\text{-} 4$ | $3$ |

$\text{-} 2$ | $2$ |

$0$ | $1$ |

$2$ | $0$ |

$4$ | $\text{-} 1$ |

Show Solution

An inverse of a function reverses its $x$- and $y$-coordinates. When a function is expressed as a table of values, finding its inverse means switching the coordinates. For example, the point $(\text{-}4,3)$ on $g$ becomes $(3,\text{-} 4)$ on $g^{\text{-} 1}.$ The following table describes $g^{\text{-} 1}(x).$

$x$ | $g^{\text{-}1}(x)$ |
---|---|

$3$ | $\text{-}4$ |

$2$ | $\text{-}2$ |

$1$ | $0$ |

$0$ | $2$ |

$\text{-}1$ | $4$ |

We can graph both $g$ and $g^{\text{-} 1}$ by marking the points from both tables on the same coordinate plane.

The graphs of a function and its inverse have a noteworthy relationship. In the coordinate plane, the function $f(x)=2x-3 \text{ and its inverse } f^{\text{-}1}(x)=\frac{x+3}{2}$ are graphed. Notice that the points of $f^{\text{-} 1}$ are the reversed points of $f.$

Because the coordinates of the points are reversed, $f^{\text{-}1}(x)$ is a reflection of $f(x)$ in the line $y=x.$

This is true for all functions and their inverses.A function is said to be invertible if its inverse is also a function. An example of an invertible function is $f(x)=0.5x-2,$ because its inverse, $f^{\text{-} 1}(x)=2x+4,$ is a linear function. Consider the function $g(x)=x^2,$ whose inverse is $\pm\sqrt{x}.$

Notice that the inverse fails the Vertical Line Test. Thus, it is not a function, which, in turn means that $g$ is not invertible. However, if the domain of $g$ is restricted to $x\geq 0,$ the inverse, $g^{\text{-} 1}(x)=\sqrt{x},$ is a function.

Thus, some functions that are not inherently invertible can be made invertible by restricting their domains.The Vertical Line Test determines if the graph of a relation is a function, or if each $x$-value corresponds with exactly one $y$-value. For a function to be invertible,
$\begin{aligned}
\text{each } y & \text{-value} \\
\text{must cor} & \text{respond to} \\
\text{exactly o} & \text{ne } x\text{-value.} \end{aligned}$
This leads to the **Horizontal** Line Test, which is performed by moving an imaginary horizontal line across the graph of a function. If this line intersects the graph more than once anywhere, the function is not invertible.

Perform test

Change function

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