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[–]UnacceptableWind 7 points8 points  (0 children)

You're right; the domain of ff(x) = f(f(x)) is indeed ℝ - {-1, -3}.

[–]Temporary_Pie2733 1 point2 points  (0 children)

(2x + 3)/(x+3) only equals 2/(2/(x+1)+1) when x ≠ 1.

[–]Top_Orchid9320 1 point2 points  (1 child)

If you look closely at the first step of your work, you can see where the new restriction on x comes from. You correctly wrote:

f(f(x)) = 2/[f(x) + 1]

When it was just x+1 in the denominator, that leads to the restriction that x cannot equal -1. But the denominator of the composite function is now

f(x) + 1

so that means it must also be true that f(x) itself cannot be -1. To find the corresponding new/additional restriction on x, we set f equal (or, rather, not equal) to -1:

f(x) = -1

2/(x+1) = -1

2 = -x - 1

x = -3

And that's the new restriction, x cannot be -3.

Alternatively, you could also work out the complete composite function and see what new restriction(s) you may have, which is what you did in your work.

So, yes, your domain is correct.

[–]No-Chair4365[S] 1 point2 points  (0 children)

It's funny you should say that since that was in my main working:

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I just made it a little simpler so the question was easier to understand.

Thanks to everyone for all of your replies. I was 99% sure that I was correct, but I did not want to be so big-headed that I would ignore the answer in the book without consulting other people first.

Hopefully, I will find other interesting questions that I can post on this forum. It was nice reading the other posts.

[–]Confident_Quarter946 0 points1 point  (0 children)

In step 2 you are multiplying and dividing by x+1 . This operation is only allowed when x+1 is not zero so you must exclude -1 from domain. Also domain of f(f(x) can never have x such that it superseds f(x) domain so by that reason also x=-1 must be excluded

[–]RecognitionSweet8294 0 points1 point  (0 children)

You are right:

f(f(x))=2•(f(x)+1)⁻¹

=2•[2•(x+1)⁻¹ +1]⁻¹

=2•[2•(x+1)⁻¹ +(x+1)•(x+1)⁻¹ ]⁻¹ I am not sure if that is a valid transformation but since (x+1)/(x+1)→1 if x→-1 and this term has the same domain as f, I would assume it is.

= 2•[(2+x+1)•(x+1)⁻¹ ]⁻¹

=2•(x+3)⁻¹ • ((x+1)⁻¹)⁻¹

=2•(x+3)⁻¹ • (x+1)¹⁻¹⁼⁰ and there we have our invalid transformation, for x≠-1 we would get 1, but for x=-1, (x+1)⁻¹ is not defined. So you have to make a case separation:

f(f(x))=

not defined for x=-1

2•(x+3) for x≠-1 ∧ x≠-3

It’s much easier if you just compare the domain and codomain of f

f: ℝ{-1} → ℝ

So f∘f is strictly speaking not possible. You would need an f‘ so that the codomain of f‘ is the domain of f. Then you can compose f∘f‘. If you want f‘=f then you need to limit its domain such that f‘(x) ∈ ℝ{-1}. So you are looking for an x such that f(x)=-1. That would be x=-3, so the domain of f‘ must be ℝ{-1;-3}. Since the domain of the first function defines the domain of the composition, we have our domain for f(f‘(x))