Question

f(x)=Sqrt{x}/(Sqrt{x}-3),find the inverse and domain and range of this function.

Answer 1

\[f(x)=\frac{\sqrt{x}}{\sqrt{x}-3}\] like this?

Answer 2

yes

Answer 3

domain.
\[x\geq 0,x\neq 9\] clear yes?

Answer 4

yes

Answer 5

range i believe is all real numbers except of course 1

Answer 6

for inverse put
\[x=\frac{\sqrt{y}}{\sqrt{y}-3}\] and solve for y

Answer 7

thats where my problem is i am having problem solving for or getting the inverse

Answer 8

\[(\sqrt{y}-3)x=\sqrt{y}\]
\[x\sqrt{y}-3x=\sqrt{y}\]
\[x\sqrt{y}-\sqrt{y}=3x\]
\[\sqrt{y}(x-1)=3x\]
\[\sqrt{y}=\frac{3x}{x-1}\]

Answer 9

Then square both sides right?

Answer 10

then square. check my algebra because it is late

Answer 11

how is range all real numbers expect 1...can u pls explain it...

Answer 12

looks good

Answer 13

i get
\[f^{-1}(x)=\frac{9x^2}{(x-1)^2}\]

Answer 14

well i think it may be a calc problem but one thing is for sure. a fraction is only when when the numerator and the denominator are the same. and that is impossible here, by inspection

Answer 15

i meant to say a fraction is only ONE when the top and bottom are equal

Answer 16

this thing has a vertical asymptote at 3, and it goes to minus infinity from one direction and positive infinity from the other. look at a picture and you will see it. but it can never be 1

Answer 17

i got vertical asymptote at 9...and picture on my calculator is at 4th quadrant only...how to get correct picture

Answer 18

what satellite meant to say is there is a vertical asymptote at x=9
not 3
he made a type-0

Answer 19

but how is range all real numbers except 1... i still cant get that..

Answer 20

i dont get it either

Answer 21

i dont get it either

Answer 22

just believe i am sure it is right

Answer 23

now you are calling them typos. see!

Answer 24

ok first of all it is clear that f cannot be 1. now you have a formula for the inverse. so what else can't f be?

Answer 25

the domain of the inverse is all real numbers except 1. that is because f cannot be 1!

Answer 26

but f can be anything else you like and your inverse proves it. suppose i want f to be 7. ok i just take
\[f^{-1}(7)\] and that finds the x that gives it to me

Answer 27

so i say the range of f is all numbers except 1, and if you give me any other number z i will take
\[f^{-1}(z)\] and say look that is the x that makes
\[f(x)=z\]

Answer 28

ohhk i got it...btw is there any way we can determine range jus by looking at the function...not its inverse..

Answer 29

i have seen people do it by "solving for x" which is basically finding the inverse. calc is easier. but you might find that the inverse involves a square root and then you have to make sure what is inside is positive. myininaya can help you with that. i just graph

Answer 30

so the range would be all numbers grater than or equal to 0 but not 9

Answer 31

?

Answer 32

hey thanks......btw can we see ur graph...

Answer 33

what is the range, is it the domain of the orignal?

Answer 34

what sis the range?

Answer 35

WHAT SI THE rNAGE?

Answer 36

so range is all numbers except 1 for the original function,also can be said as domain of the inverse

Answer 37

but what is range of inverse?

Answer 38

range of inverse is domain of the original function

Answer 39

all real numbers except 9

Answer 40

hey man do you get related reates

Answer 41

still on quiz qs....

Answer 42

but it cant be all reall numbers, it has to numbers greater than o equal to zero except 9

Answer 43

havent got that far...

Answer 44

i am done with it

Answer 45

look at this post and click on it, i will give them to you

Answer 46

http://www.twiddla.com/582353