Determing if an equation is a function

[Ry122]

How do i determine if ((x-4)^(1/2))+3=y is a function? How do I determine the domain and range?

 

[rock.freak667]

Well by the definition of a function...that would not be considered a function as you can find one value of x that corresponds to two values of y.

to make that equation a function, you must specify whether the +ve or -ve of the square root must be taken

 

[sprint]

for every number you plug into X, there will be a number Y that would result.

so yes, it is a function.

domain is all positive numbers since you can't take the square root of a negative number.

well you can but I am assuming your teacher doesn't grade at that kind of level.

and again I am making assumptions because of the nature of your question.

 

[danago]

for every number you plug into X, there will be a number Y that would result.

so yes, it is a function.

domain is all positive numbers since you can't take the square root of a negative number.

well you can but I am assuming your teacher doesn't grade at that kind of level.

and again I am making assumptions because of the nature of your question.

What about x=1? 1 is a positive number, and it still results in taking the square root of a negative number.

 

[Ry122]

Since ((x-4)^(1/2))+3 is the same as sqroot(x-4)+3 doesn't this mean that
x-4=-3^2
(x-4)=9
and therefore it is a function?
In my book however it says that it is not a function and the domain is x > 4 andrange is
y is an element of IR

 

[CompuChip]

That doesn't make sense, the book says it is not a function, but still talks about the domain and range?

You have a prescription where you can plug in a number x, and get a unique number y (or none at all, but not multiple), so it is a function. To find the domain, you correctly remarked that you cannot take the square root of a negative number. So you should check for which x you will not get a negative number under the square root sign. But note, that to talk about the domain, you first have to establish that it is a function, it's no use doing a calculation and then saying: look I found a domain, therefore it is a function.

 

[sprint]

What about x=1? 1 is a positive number, and it still results in taking the square root of a negative number.

yeah you are right. i meant {X: (X > 4) U (X =4)}

 

[CompuChip]

Careful with the notation!
You mean,
{X : (X > 4)} U {4}
or if you insist on writing it stupidly,
{X : (X > 4)} U {X: X = 4}
or, as it would be commonly written,
\{X : X \ge 4 \}

 

[sprint]

whoa... don't need to throw insults around einstein

i would have written ex greater than or equal to four but i don't know how to use all that fancy latex stuff

besides, there was no error in writing it the way i wrote the domain. it is too anal to declare the variable twice... or to even have this conversation.

 

[CompuChip]

Whoa, didn't know you were going to take that as an insult. I just wanted to point it out to you, because when people (in general) start writing domains and ranges, they often make mistakes in the notation and IMO it's important that one learns to write it correctly. So my remark was merely meant as constructive criticism. My apologies if you misunderstood my intentions.

By the way, if you can't use LaTeX, you can always use >= and anyone will understand what you mean (at least, I suppose anyone on this forum will) .