+0  
 
0
47
6
avatar

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?)

 
 Oct 11, 2021
edited by Guest  Oct 11, 2021
 #1
avatar+114826 
+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.}\\
 

 
 Oct 12, 2021
 #2
avatar
+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
avatar+114826 
+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

 

https://www.desmos.com/calculator/ph3tahg9ut

 
Melody  Oct 12, 2021
 #4
avatar
+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
avatar+114826 
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 wink

 
Melody  Oct 12, 2021
edited by Melody  Oct 12, 2021
 #6
avatar
+1

Maybe ! (will consider it). But, thanks for the help!  smiley

 
Guest Oct 12, 2021

32 Online Users

avatar
avatar
avatar