Question

How to find the domain and range of a composite function?

Answer 1

really it depends on the functions.

Answer 2

Let say I have f(x)=1/(2-x) and I need to find f.f(x).

Answer 3

first of all to be in the domain of the composite function it must first be in the domain of the "inside " funciton

Answer 4

The domain cannot be 2 and 3/2 but how do I find the range then.

Answer 5

and secondly the out put must be in the domain of the "outside function"

Answer 6

you have
\[f(x)=\frac{1}{2-x}\] right? so right away the domain is all numbers except
\[x=2\]

Answer 7

The function is f(f(x)).

Answer 8

then you want
\[f\circ f(x)=f(f(x))\] right?

Answer 9

Yup.

Answer 10

so that is
\[f(f(x))=f(\frac{1}{2-x})\]
\[=\frac{1}{2-\frac{1}{2-x}}\]

Answer 11

Yes.

Answer 12

and if my algebra is correct, then this is
\[f\circ f(x)=\frac{2-x}{3-2x}\] right?

Answer 13

Yup that's what I have as well.

Answer 14

ok then the domain of
\[f\circ f\] is all numbers except
\[2,\frac{3}{2}\]

Answer 15

Yup. But I don't know how to find the range.

Answer 16

well that is a much much harder question

Answer 17

I saw my teacher plugging those values in something.

Answer 18

Could it be the inverse?

Answer 19

without calculus it is not straightforward, but i can tell you how i know pretty easily first of all this is the same as
\[\frac{x-2}{2x-3}\]

Answer 20

and the range will be all numbers except
\[\frac{1}{2}\]

Answer 21

Did you differentiate the fcn?

Answer 22

that is from common sense. this ratio cannot be 1/2 because the numerator would have to be half of the denominator, and that is not possible because whatever x is you will have that -2 on top and the -3 on the bottom

Answer 23

another way to see it is that the "horizontal asymptote" is
\[y=\frac{1}{2}\] and this particular function does not cross its asymptote

Answer 24

but how you can tell that without calc is beyond me . i mean the asymptote is easy, but i can't think of a non - calculus way to show that it does not cross the asymptote

Answer 25

Oh, I forgot. I'm taking a calculus course right now.

Answer 26

I think finding the limits will help find the range.

Answer 27

well then take the derivative and discover that the function is alway increasing.

Answer 28

Is that correct?

Answer 29

then say that the limit as you go to infinity is 1/2 and the limit as you go to - infinity is also 1/2 so it never crosses the line y = 1/2

Answer 30

Ok, thanks. I had solve this analytically.

Answer 31

yw