The functions f and g are defined as follows:

\(f(x)=x^2-1\) for \(x<0\)

\(g(x)=\dfrac{1}{2x+1}\) for \(x<-\frac{1}{2}\)

Find the domain and range of \(f(g(x))\), and the domain and range of \((fg)^{-1}(x)\)

So,I found that \(f(g(x))=\frac{1}{(2x+1)^2}-1\)

And since, the range of g(x) is the domain of f(x) (otherwise, the composition won't work) then the domain of \(f(g(x))\) is \(x<-\frac{1}{2}\)

And hence, the range is, \(f(g(x))>-1\) (Subsituiting large negative value of x, results in fg(x) approaching -1, so it must always be greater than -1, and as x gets closer to -1/2, it approaches infinity)

Is this working correct so far? Please help; thanks!

(the domain and range of \((fg)^{-1}(x)\) is the reverse of domain and range of \(f(g(x))\) right?)

Guest Oct 11, 2021

edited by
Guest
Oct 11, 2021

#1**+1 **

Mmm I am not great at these either but here is my logic.

\(f(x)=x^2-1\;\;for\;\;x<0\\ g(x)=\frac{1}{2x+1}\;\;for \;\;x<-0.5\\~\\ f(g(x))=(\frac{1}{2x+1})^2+1\qquad \text{domain to be determined}\\~ \\for\\ f(g(x)) \quad \text{Firstly } x <-0.5\\ but \;\;also\\ \frac{1}{2x+1}<0\\ \text{Yes that must already be true, no more restrictions necessary}\\ \text{So the domain will be }x<-0.5\\ \text{Now I will look at the range.}\\ \)

As x tends to -0.5 from below f(g(x)) tends to infinity

As x tends to -infinity f(g(x)) tends to -1

So the range is (-1, +infinity)

**I think that is the same as you got. **

And yes your last statement is correct.

But it was necessary to think about whether any further restraints were necessary for the inverse to still be a function. (no restraints are necessary)

Here are the graphs to back up our findings.

https://www.desmos.com/calculator/o0avvsjz6h

Some of your logic seemed simpler than mine but I am not sure if it was entirely trustworthy.

LaTex

f(x)=x^2-1\;\;for\;\;x<0\\

g(x)=\frac{1}{2x+1}\;\;for \;\;x<-0.5\\~\\

f(g(x))=(\frac{1}{2x+1})^2+1\qquad \text{domain to be determined}\\~

\\for\\

f(g(x)) \quad \text{Firstly } x <-0.5\\

but \;\;also\\

\frac{1}{2x+1}<0\\

\text{Yes that must already be true, no more restrictions necessary}\\

\text{So the domain will be }x<-0.5\\

\text{Now I will look at the range.}\\

Melody Oct 12, 2021

#2**+1 **

Thank you very much, that was helpful (especially the graph).

I usually apply the following rule: (Read it online.)

Consider the composite function f(g(x))

-->The domain of f(g(x)) is the domain of g since that is the function which operates on the variable x and the range of f(g(x) is the range of f since that is the range of the output.

Guest Oct 12, 2021

#3**+1 **

I think that is too simplistic:

Consider this new example.

f(x)=3x where x>4

g(x)=x^2 where x<-1

f(g(x))=3x^2

certainly x must be less than -1

but

x^2 must also be bigger than 4 which means that x>2 or x<-2

The intersection of these two restrictions, x<-1 and x<-2 is x<-2

so I believe that the domain of f(g(x)) is x<-2

Melody
Oct 12, 2021

#4**+1 **

Oh! Right!

Now, I think the best method is to find the common intersection of the domains as you did.

Guest Oct 12, 2021

edited by
Guest
Oct 12, 2021

#5**0 **

As I said, I am not ultra-confident with this topic but that sounds right to me.

Why don't you become a member?

It is easier to track your posts.

Members do tend to get priority.

and it is easier to attract people's attention through private messaging.

Plus it is completely free.

I think you can also get email alerts when you get a question response too (although I am not certain on that point)

You could even make some good friends here. I have

Melody
Oct 12, 2021