{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} For the given function, we will first draw its graph and then use the Horizontal Line Test to determine whether the inverse is a function. If it is, then we will find the inverse. Let's start!

To graph the given function, we should first determine its asymptotes. $f(x)=11−2x4 $ Recall that division by zero is not defined. Therefore, the rational function is undefined where $11−2x=0.$ $11−2x=0⇔x=5.5 $ At $x=5.5$ the function has a vertical asymptote. To find the horizontal asymptote, let's pay close attention to the degrees of the numerator and denominator. $f(x)=11−2x_{1}4 $ We see that the degree of the denominator is higher than the degree of the numerator. Therefore, the line $y=0$ is a horizontal asymptote. Now, we will draw the asymptotes.

Next, we will make a table of values. We will include $x-$values to the left and to the right of the vertical asymptote.

$x$ | $11−2x4 $ | $f(x)$ |
---|---|---|

$2$ | $11−2(2)4 $ | $≈0.57$ |

$3$ | $11−2(3)4 $ | $0.8$ |

$4$ | $11−2(4)4 $ | $≈1.33$ |

$5$ | $11−2(5)4 $ | $4$ |

$6$ | $11−2(6)4 $ | $-4$ |

$7$ | $11−2(7)4 $ | $≈-1.33$ |

$8$ | $11−2(8)4 $ | $-0.8$ |

$9$ | $11−2(9)4 $ | $≈-0.57$ |

Finally, let's plot and connect the points.

Now, we can apply the Horizontal Line Test.

We can see above that there is no horizontal line that intersects the graph at two or more points. Therefore, the inverse of the given function is also a function.

$x=11−2y4 $

Solve for $y$

MultEqn$LHS⋅(11−2y)=RHS⋅(11−2y)$

$x(11−2y)=4$

DivEqn$LHS/x=RHS/x$

$11−2y=x4 $

AddEqn$LHS+2y=RHS+2y$

$11=x4 +2y$

SubEqn$LHS−x4 =RHS−x4 $

$11−x4 =2y$

NumberToFrac$a=xx⋅a $

$x11x −x4 =2y$

SubFracSubtract fractions

$x11x−4 =2y$

DivEqn$LHS/2=RHS/2$

$2x11x−4 =y$

RearrangeEqnRearrange equation

$y=2x11x−4 $