1. ## Function

Hey Guys I've got a slight problem regarding the following question. Im enrolling into a course next year and im having problems solving the practice examples.

[IMG]file:///C:/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]

2. ## Re: Function

Originally Posted by mikel03
Hey Guys I've got a slight problem regarding the following question. Im enrolling into a course next year and im having problems solving the practice examples.

[IMG]file:///C:/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
And what are the problems you are having?

4. ## Re: Function

Sorry but that's not good enough, you need to have at least tried something... Do you understand what a function is (inputting numbers into a rule and getting a number out)? Do you understand that the domain of a function is a list of all the possible values you can put in (in other words, all the possible x values) and that the range is a list of all the possible values you can get out (in other words, all the possible y values)?

5. ## Re: Function

ive got this but im unsure if im correct or along the right track.

y1 domain -3<x<3. range all positive real numbers<=3
x=+-3 asymptotes

y2 domain all real numbers not = 2
range all real numbers
x=2 asymptote

6. ## Re: Function

Originally Posted by mikel03
ive got this but im unsure if im correct or along the right track.

y1 domain -3<x<3. range all positive real numbers<=3
x=+-3 asymptotes

y2 domain all real numbers not = 2
range all real numbers
x=2 asymptote
y1 domain is correct. You are also correct that x = -3 and x = 3 are vertical asymptotes. The range is incorrect. Have you tried looking at a graph of the function?

y2 domain is correct and x = 2 is an asymptote. The range is incorrect. I suggest you try long dividing to get the function into a form that is easier to analyse...

7. ## Re: Function

Thanks for your help but im unsure ive tried multiple times to solve the range and i got the same asnwer.

can you please identify what im doing wrong

8. ## Re: Function

Originally Posted by mikel03
Thanks for your help but im unsure ive tried multiple times to solve the range and i got the same asnwer.

can you please identify what im doing wrong
$\displaystyle \displaystyle \frac{1}{\sqrt{9 - x^2}}$ will be at its minimum where $\displaystyle \displaystyle \sqrt{9 - x^2}$ is at its maximum. You should recognise this as the upper half of a circle, what is its maximum value?