$$f(x) = 2 - |x|, -1\le x \le 3 \text{ and} g(x) = |x + 1| -1, -2\le x \le 2\text{. Determine $f\{g(x)\}$}$$

Question

$$f(x) = 2  - |x|, -1\le x \le 3 \text{ and} g(x)  = |x + 1| -1, -2\le x \le 2\text{. Determine $f\{g(x)\}$}$$

anonymous · Answer

Determine f[g(x)]

anonymous · Answer

Do I have to include g(x)'s domain too

anonymous · Answer

Hmm It's -2<x<2 in the book

zarkon · Answer

you don't intersect when it is composition of functions

anonymous · Answer

I think it's -1<f(x)<3 and -2<x<2

myininaya · Answer

why wouldn't f o g have domain where f and g both have domain

anonymous · Answer

if I solve -1<f(x)<3 
i am getting -5<x<3 and it's intersection with -2<x<2 gives -2<x<2 as the answer which is the answer in the book but the thing that worries me is that I have been ignoring g(x)'s domain (fogox) until now :-/

zarkon · Answer

$$f:C\to D$$
$$g:A\to B$$
then to have $$f(g(x))$$ we just need $$B\subseteq C$$

then the domain of $f(g(x))$ is A

myininaya · Answer

ok zarkon sounds good

myininaya · Answer

teach me algebra

anonymous · Answer

Cool Zarkon that is really going to help me thanks

myininaya · Answer

hes so smart

anonymous · Answer

Indeed 

that was awesome

myininaya · Answer

he probably could have got his phd in kindergarten

zarkon · Answer

using my notation...if B is not a subset of C then you have work to do :)

anonymous · Answer

put gx in domain of fx,

and solve by breaking gx into appropriate domains,

anonymous · Answer

hello there，if B is not a subset of c ,then there is no answer

anonymous · Answer

Stom!!! have a look at Zarkon's solution

zarkon · Answer

yes there could be

anonymous · Answer

how?

anonymous · Answer

@zarkon domain of fgx should be the range of gx,
hence B and not A

zarkon · Answer

$f(x)=x$ on [-1 to 1]
$g(x)=\sqrt{x}$ on [0,2]

then g(f(x)) is defined on [0,1]

zarkon · Answer

using my notation the domain is A

anonymous · Answer

domain of fgx should be the range of gx

zarkon · Answer

no...it is not

anonymous · Answer

yes it is

zarkon · Answer

...think about it

anonymous · Answer

i did, and even checkeed notes

zarkon · Answer

f(g(x))

you first plug in values into g...g has domain A...

anonymous · Answer

i think about this like this,

fgx will take those values which will be given by gx, hence range of gx,

gx returns values which gets into fgx,

hence domain fgx is range of gx

anonymous · Answer

Stom look at Zarkon's solution and solve some problems on composite functions it is true I just tried some problems

zarkon · Answer

$$g:A\to B$$
$$f:B\to C$$

$$f\circ g:A\to C$$

zarkon · Answer

"gx returns values which gets into fgx,"

No...using your wording ...  gx returns values which gets into fx

anonymous · Answer

g(x) is range here is  [-1, 2] but the domain for f(g(x)) is [-2,2]
g(x) = |x + 1| - 1

anonymous · Answer

f: a to b

g: c to d

fog: d to e

whats wrong with this

zarkon · Answer

g might not be defined on d