Question

What is wrong with this Domain and Range?

y = -2x -5

{xeI / x<-15/2 , x>-2} {yeI / y>10 , y<-1}

Answer 1

Er.. the domain is \(x\in \mathbb{R}\) as is the range?

Answer 2

the range was the restriction that the teacher gave us and it started with \[y \epsilon \]

Answer 3

I mean y is an element of I

Answer 4

All right, using that, then, for some domain:
\[
x\in(a,b)
\]We must have a range of a linear function of:
\[
f(x)\in\left(\;\min(f(a), f(b)),\; \max(f(a), f(b)\right)\;)
\]

Answer 5

wait what does the I even mean?

Answer 6

I don't know... l? That's not stated in the problem...?

Answer 7

I'm not sure if it means y is an element of integers or something like that... but my teacher said my calculations were right except that I got something fundamentally wrong with my domain...

Answer 8

No, the integers is \(\mathbb{Z}\), and your region is correct. Just that, you cannot say that \(x,y\in \mathbb{I}\), because \(\mathbb{I}\) typically stands for an identity function, not the integers. Instead, you should say: \(x,y\in \mathbb{R}\), which means reals.

Answer 9

so for my domain it should be x∈R ?

Answer 10

OH. I think I get what your question is. So you were given the constrains for \(y\) to be an integer and had to find the fitting \(x\)? Am I incorrect? In such a case, yes there is something fundamentally wrong in your answer, as \(-\frac{15}{2}\) is not an integer.

Answer 11

yes the range was the constraint and I had to find the domain. So does y∈I mean that I need to work with integers only?

Answer 12

so do I have to round it?

Answer 13

\(y\in \mathbb{I}\) does not exist. \(\mathbb{I}\) is not a real field.

Answer 14

soo what does that mean for the domain LOL

Answer 15

there is no solution?