# Thread: Domain and range - functions and relations :D

1. ## Domain and range - functions and relations :D

Hello all!

We're currently learning about Functions and Relations. Here is a question im unsure about. Help would be greatly appreciated.

Question: For the following function, state:

(i) the domain
(ii) the co-domain
(iii) the range
f : ( -¥ , 0 ) ® R , f (x) = 1/ Ö- x (one divided by square root of negative x)

I'm more concerned about how to do the range... because the answer says that the range is: ( 0 , ¥ ). How do you know with those types of graphs ( sqaure root underneath ) what the range is?

Help would be greatly appreciated! Thank- you

2. Originally Posted by steph_r
Hello all!

We're currently learning about Functions and Relations. Here is a question im unsure about. Help would be greatly appreciated.

Question: For the following function, state:

(i) the domain
(ii) the co-domain
(iii) the range

f : ( -¥ , 0 ) ® R , f (x) = 1/ Ö- x (one divided by square root of negative x)

I'm more concerned about how to do the range... because the answer says that the range is: ( 0 , ¥ ). How do you know with those types of graphs ( sqaure root underneath ) what the range is?

Help would be greatly appreciated! Thank- you
If you have

$\displaystyle f(x)=\frac1{\sqrt{-x}}$

Then you have to consider: 1. the radical must be greater or equal zero and 2. the denominator must be unequal zero.

to 1.: $\displaystyle -x\geq 0~\implies~ x \leq 0$

to 2.: $\displaystyle x \ne 0$

therefore the domain of your function is $\displaystyle x \in (-\infty, 0)$

Since $\displaystyle \sqrt{-x} > 0$ the range of the function are all positive numbers:

$\displaystyle y \in (0, +\infty)$

3. ## Functions

So.. does the line of that equation, when graphed on the calculator, actually touch the y-intercept? because im unsure about why the ¥ is there .

Thank you

4. Originally Posted by steph_r
So.. does the line of that equation, when graphed on the calculator, actually touch the y-intercept? because im unsure about why the ¥ is there .

Thank you
As I've posted in my previous post zero doesn't belong to the range. To illustrate the behaviour(?) of the function I've attached the graph of f: