Piecewise continuity problem

[coooool222]

These piecewise functions are bugging me. This piecewise function is continuous everywhere except for 5. But my answer key is telling me that it's also not continuous in -6 < x =< -1. I don't see that interval anywhere in the problem. I
also do not get the addition of the x =< -6.

I am not allowed to graph this piecewise function.

 

[BigBeachBanana]

View attachment 34722
These piecewise functions are bugging me. This piecewise function is continuous everywhere except for 5. But my answer key is telling me that it's also not continuous in -6 < x =< -1. I don't see that interval anywhere in the problem. I
also do not get the addition of the x =< -6.

I am not allowed to graph this piecewise function.

What's the value of [imath]f(-1), f(-2) = ?[/imath]
The function is not defined [imath]-6 < x \le -1[/imath].
You don't necessarily need to draw a graph. Drawing a simple number line to visualize the domain of the function is sufficient.

 

[coooool222]

What's the value of [imath]f(-1), f(-2) = ?[/imath]
The function is not defined [imath]-6 < x \le -1[/imath].
You don't necessarily need to draw a graph. Drawing a simple number line to visualize the domain of the function is sufficient.

do i always have to draw a number line?

 

[BigBeachBanana]

do i always have to draw a number line?

It's not required, but it helps if you can't see it.

 

[Otis]

do i always have to draw a number line?

Hi. My experience says, "yes – except when you don't need to."
[imath]\;[/imath]

 

[Steven G]

Draw any graph that you like in the given intervals (or use a number line) and see where the function is undefined.

 

[pka]

View attachment 34722
These piecewise functions are bugging me. This piecewise function is continuous everywhere except for 5. But my answer key is telling me that it's also not continuous in -6 < x =< -1. I don't see that interval anywhere in the problem. I
also do not get the addition of the x =< -6.

I am not allowed to graph this piecewise function.

The set [imath]\mathcal{D}=(-\infty,-6] \cup(-1,2)\cup [2,\infty)[/imath] is the domain of the function [imath] y(x)[/imath] below.

[imath]y(x)=\begin{Bmatrix} \dfrac{1}{x-5}& x\ge 2\\ \dfrac{-1}{3} &-1<x< 2 \\ 3x & -x\le -6 \end{Bmatrix}[/imath] Now note that: [imath]\mathop {\lim }\limits_{x \to 2} y(x) = \dfrac{{ - 1}}{3}[/imath].

BUT [imath]\bf\color{red}\mathop {\lim }\limits_{x \to -1}y(x) ~\&~\mathop {\lim }\limits_{x \to -6}y(x) [/imath] do not exist because of domain issuers. In order for the function to be continuous at a point it is necessary for the function to be defined at the point and the limits from the left and from the right must exist an be equal.

 

[stapel]

View attachment 34722
...my answer key is telling me that it's also not continuous in -6 < x =< -1. I don't see that interval anywhere in the problem.

How can a function be continuous (or have any other properties) where it doesn't even exist?

Eliz.