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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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, $f_{-1},y$ is the input and $x$ would be the corresponding output. $f(x)=y⇔f_{-1}(y)=x$

Some function families are inverses of each other. This is because some functions undo each other. For example, $x_{2}$ and $±x $ are inverses because radicals and exponents (with the same index) undo each other.Some of the coordinates of the function $g$ are shown in the table. Find $g_{-1},$ then graph $g$ and $g_{-1}$ on the same coordinate plane.

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

$-4$ | $3$ |

$-2$ | $2$ |

$0$ | $1$ |

$2$ | $0$ |

$4$ | $-1$ |

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 $(-4,3)$ on $g$ becomes $(3,-4)$ on $g_{-1}.$ The following table describes $g_{-1}(x).$

$x$ | $g_{-1}(x)$ |
---|---|

$3$ | $-4$ |

$2$ | $-2$ |

$1$ | $0$ |

$0$ | $2$ |

$-1$ | $4$ |

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

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 $f(x)=32x−1 .$
### 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)=32x−1 ⇔y=32x−1 $

### 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. $y=32x−1 ⇒x=32y−1 $

### 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=32y−1 $

MultEqn$LHS⋅3=RHS⋅3$

$3x=2y−1$

AddEqn$LHS+1=RHS+1$

$3x+1=2y$

DivEqn$LHS/2=RHS/2$

$23x+1 =y$

RearrangeEqnRearrange equation

$y=23x+1 $

Replace $y$ with $f_{-1}(x)$

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

$y=23x+1 ⇔f_{-1}=23x+1 $

Notice that in $f,$ the input is multiplied by $2,$ decreased by $1$ and divided by $3.$ From the rule of $f_{-1},$ 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 $2.$

Consider the quadratic function $f(x)=3x_{2}.$ Find its inverse function for when $x>0.$

Show Solution

To find the inverse of $f(x),$ we first need to replace $f(x)$ with $y.$
$f(x)=3x_{2}⇒y=3x_{2}$
The next step is to switch $x$ and $y$ in the function rule. $y=3x_{2}⟶switch x=3y_{2}$
Now we need to solve for $y.$ The resulting equation will be the inverse of the given function.
Switching $x$ and $y$ also switches the domain and the range. This means that the restriction on the domain of the function, $x>0,$ becomes a restriction on the range of the inverse function, $y>0.$ Thus, there are no negative $y$-values. $y=3x $
Finally, to indicate that this is the inverse of $f(x),$ we will replace $y$ with $f_{-1}(x).$ $y=3x ⇒f_{-1}(x)=3x $

$x=3y_{2}$

DivEqn$LHS/3=RHS/3$

$3x =y_{2}$

RearrangeEqnRearrange equation

$y_{2}=3x $

SqrtEqn$LHS =RHS $

$y=±3x $

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