# Domain of f o g

1. Sep 15, 2007

### rootX

1. The problem statement, all variables and given/known data

if
$$f\left( x\right) :=\frac{1}{x-1}$$
for f(x) Doman (x != 1)
$$g\left( x\right) :=\frac{1}{\sqrt{x+1}}$$
for g(x), Domain (x >= -1)

so

for f(g(x))
the Domain would be (-1,0) U (0, inf)?

or do I need to consider other things?

2. Sep 15, 2007

### rootX

I was also wondering in composite functions,
if I find domain after simplifying all the composite function, would that be correct?

Or I need to find domain restrictions for f(x), and then on g(x)...

3. Sep 15, 2007

### dextercioby

The domain is in words: all "x" in the domain of g such as the range of g is in the domain of f. And the answer is indeed (-1,0) reunion (0, +infty).

4. Sep 15, 2007

### rootX

so, do I always need to go that way
find domain of g(x),

and any restrictions would be restrictions for f o g

and then find domain of f(x)
and any restriction on that would be restriction for the range of g(x)..

and then domain of f o g = domain of g U (range of g(x) and domain of f(x))..

or, simply finding domain of f o g from the f of g function - the one I get once I subsitute the g(x) in it?

Q2/ for inverse functions why we reflect them about y = x func?

5. Sep 15, 2007

### rootX

used vectors to the answer to # 2..

first dirc vector between (x,y) and (y,x) is (1,-1)

and thus norm to y =x

and then distance is always same

(x-a)^2 + (y-a)^2 = (y-a)^2 + (x-a)^2!!
@@

6. Sep 15, 2007

### learningphysics

Yes, but don't simplify the equation once you do the substitution... ie: f(g(x)) = $$\frac{1}{\frac{1}{\sqrt{x+1}}-1}$$

so looking at this function... what's the domain?

7. Sep 15, 2007

### learningphysics

looks good to me!

8. Sep 15, 2007

### rootX

I get the same answer, but simplification also gives same domain.

but I think it would be tedious if they ask for the domain of

f(g(h(....k(x).....)))

9. Oct 31, 2008

### roam

I also have a relevent/similar question but a bit confusing. We have two functions: f(x) = $$\sqrt{1+x}$$ and g(x) = $$\frac{x}{x-1}$$

We want to find the domain of f o g.

D(f) => 1+x2 $$\geq$$ -1

D(f) = [-1, $$\infty$$)

D(g) = R\{1}

Now, the domain of f o g is: D(g) such that the g(x) $$\in$$ D(f).

=> $$\frac{x}{x-1} \geq -1$$

$$\frac{2x-1}{x-1} \geq 0$$

Two points are 1, $$\frac{1}{2}$$. Testing between these;

x<1/2 => +ve
1/2<x<1 => -ve
x>1 => +ve

Therefore the domain of f o g must be $$(-\infty, \frac{1}{2})$$ U $$(1, \infty)$$

The correct answer has to be: D(f o g) = [1/2, 1]. Can anyone explain this for me please?

Last edited: Oct 31, 2008
10. Oct 31, 2008

### HallsofIvy

Staff Emeritus
Why does the domain "have to be" [1/2, 1]? If x= 3/4, which is in that set, then g(3/4)= (3/4)/(-1/4)= (3/4)(-4)= -3 and f(-3) is not defined. That is essentially the analysis you did. Obviously f(g(x)) is NOT defined at 3/4 and so not defined on [1/2, 1].

11. Oct 31, 2008

### roam

Yes, I understand that but I checked the answers at the back of the book and it says D(f o g) = [1/2, 1]! That's what makes me confused.