Dental Plan
Badges: 0
Rep:
?
#1
Report Thread starter 7 years ago
#1
Hi guys, hope you can help.

I'm self-teaching Maths A level this year and currently following the "Advancing Maths for AQA" series. I'm a little stuck on answering the following question, since the book hasn't told me how (AFAIK).

Find the range of the function g(x) = \frac{5x-4}{2x+1}, x \in \mathbb{R}, x \not= -\frac{1}{2}

Now, I know the range is the set of values the function can take for the domain, but is there a quick way to work it out here?

My only idea is to find the domain of the inverse function, \frac{x+4}{5-2x}, which I see clearly see is x \not= \frac{5}{2} (since we cannot divide by zero).

Am I overlooking an easier way to find this solution?

Many thanks in advance,

Dental plan (Lisa needs braces)
0
reply
ghostwalker
  • Study Helper
Badges: 17
#2
Report 7 years ago
#2
(Original post by Dental Plan)

Find the range of the function g(x) = \frac{5x-4}{2x+1}, x \in \mathbb{R}, x \not= -\frac{1}{2}
The range is the set of values in the codomain.

Your method is as good as any.

You could rewrite g(x) as \frac{5}{2}-\frac{6.5}{2x+1}

Hopefully it's clear that this is a transformation of 1/x, whose range is all values except 0.

In this question the fraction can be anything but zero, so the range is anything, except 5/2.
0
reply
Dental Plan
Badges: 0
Rep:
?
#3
Report Thread starter 7 years ago
#3
Thanks for your reply, but I'm still struggling. I'm not quite sure what you mean by a "transformation of 1/x" (I've only just started C3 and pretty bad at maths generally). If you had a chance, could you explain how you rewrite g(x) like this? Would be great.

And the reason I don't want to use my method of finding the domain of the inverse function was just that the question after this one asked for the inverse function, which seemed to imply I could find the range to the first one without finding the inverse first.

Thanks again.
0
reply
WhiteGroupMaths
Badges: 9
Rep:
?
#4
Report 7 years ago
#4
(Original post by Dental Plan)
Hi guys, hope you can help.

I'm self-teaching Maths A level this year and currently following the "Advancing Maths for AQA" series. I'm a little stuck on answering the following question, since the book hasn't told me how (AFAIK).

Find the range of the function g(x) = \frac{5x-4}{2x+1}, x \in \mathbb{R}, x \not= -\frac{1}{2}


You can consider sketching the graph of g(x) = \frac{5x-4}{2x+1}, x \in \mathbb{R}, x \not= -\frac{1}{2}, and examine the feasible set of y values corresponding to the given domain. Should take you no longer than 30 sec.

As far as graph sketching is concerned, you may wish to take a look here:

http://www.msharpener.com/2012/07/quick-graphing-2.html

Hope it helps. Peace.
0
reply
ghostwalker
  • Study Helper
Badges: 17
#5
Report 7 years ago
#5
(Original post by Dental Plan)
Thanks for your reply, but I'm still struggling. I'm not quite sure what you mean by a "transformation of 1/x" (I've only just started C3 and pretty bad at maths generally). If you had a chance, could you explain how you rewrite g(x) like this? Would be great.
Rewriting g(x) is via long/polynomial division, which is only relevant if you're already covered it.

And the actual graph transformation are not really relevant. I mentioned 1/x as you should know the shape of that function, and that it's range includes all reals except zero. Transforming it to the new fraction you have stretches x,y and shifts in the x direction, but does not shift in the y direction, so the value it can't be is still zero. Hence the whole thing can't be 5/2.

Alternatively, look at your original fraction and divide top and bottom by x, to get (5-[4/x])/(2+[1/x])

We can see that as x goes to infinity, your fraction goes to 5/2 (the terms in "thing/x" go to zero) but never actually reaches it.

Since it's decades since I did A-level, the things I'm suggesting to you, you may not have covered at this stage, C3. Ignore the ones you haven't done.
0
reply
L'Evil Fish
Badges: 18
Rep:
?
#6
Report 7 years ago
#6
Can I ask....

When combining functions, what are the rules with regards to domains and ranges?
1
reply
ghostwalker
  • Study Helper
Badges: 17
#7
Report 7 years ago
#7
(Original post by L'Evil Fish)
Can I ask....

When combining functions, what are the rules with regards to domains and ranges?
I presume you're not talking about function of a function, where the domain of one is the range of the other.

(OP's question was a simple case, just dealing with the standard simple transformations.)

That aside, I'd think you'd need to treat each case separately, rather than there being any rules.

E.g Knowing the domain and range of sin and cos, still requires quite a bit of work to find the range of sin + cos,
0
reply
L'Evil Fish
Badges: 18
Rep:
?
#8
Report 7 years ago
#8
(Original post by ghostwalker)
I presume you're not talking about function of a function, where the domain of one is the range of the other.

(OP's question was a simply case, just dealing with the standard simple transformations.)

That aside, I'd think you'd need to treat each case separately, rather than there being any rules.

E.g Knowing the domain and range of sin and cos, still requires quite a bit of work to find the range of sin + cos,
Hmmmm, are there any restrictions?

Like in fg(x) can the domain not exceed the domain of g(x) or?
0
reply
ghostwalker
  • Study Helper
Badges: 17
#9
Report 7 years ago
#9
(Original post by L'Evil Fish)
Hmmmm, are there any restrictions?

Like in fg(x) can the domain not exceed the domain of g(x) or?
OK, we are talking about functions of functions.

That's correct, the domain of fg, must be a subset of the domain of g, since fg(x) = f(g(x)) and g(x) wouldn't otherwise be defined.

Consider the two function:

f:A\to B\\g:C\to D

Where sets A,B,C,D are the domains and codomains.

Then the function fg is only valid where the image of g is in the domain of f.

I.e we need g(C)\cap A

So, the domain of fg is g^{-1}(g(C)\cap A)

Where g^{-1}(X) is not the inverse of g, but rather the set of elements in C that map to the set X.

Edit:

If you desire, have a go with g(x) = x-1 defined on all R, and f(x) = sqrt(x) defined on x>=0

Edit2:

It's usually probably easiest to combine the two functions into one, work out the possible domain and intersect it with C.
0
reply
maggiehodgson
Badges: 14
Rep:
?
#10
Report 7 years ago
#10
(Original post by Dental Plan)
Hi guys, hope you can help.

I'm self-teaching Maths A level this year and currently following the "Advancing Maths for AQA" series. I'm a little stuck on answering the following question, since the book hasn't told me how (AFAIK).

Find the range of the function g(x) = \frac{5x-4}{2x+1}, x \in \mathbb{R}, x \not= -\frac{1}{2}

Now, I know the range is the set of values the function can take for the domain, but is there a quick way to work it out here?

My only idea is to find the domain of the inverse function, \frac{x+4}{5-2x}, which I see clearly see is x \not= \frac{5}{2} (since we cannot divide by zero).

Am I overlooking an easier way to find this solution?

Many thanks in advance,

Dental plan (Lisa needs braces)
Thanks for that - I've struggled with these too but finding the inverse's domain just might crack it for me.
0
reply
L'Evil Fish
Badges: 18
Rep:
?
#11
Report 7 years ago
#11
(Original post by ghostwalker)
OK, we are talking about functions of functions.

That's correct, the domain of fg, must be a subset of the domain of g, since fg(x) = f(g(x)) and g(x) wouldn't otherwise be defined.

Consider the two function:

f:A\to B\\g:C\to D

Where sets A,B,C,D are the domains and codomains.

Then the function fg is only valid where the image of g is in the domain of f.

I.e we need g(C)\cap A

So, the domain of fg is g^{-1}(g(C)\cap A)

Where g^{-1}(X) is not the inverse of g, but rather the set of elements in C that map to the set X.

Edit:

If you desire, have a go with g(x) = x-1 defined on all R, and f(x) = sqrt(x) defined on x>=0

Edit2:

It's usually probably easiest to combine the two functions into one, work out the possible domain and intersect it with C.
:lol: feeling stupid now...

Don't understand.
0
reply
ghostwalker
  • Study Helper
Badges: 17
#12
Report 7 years ago
#12
(Original post by L'Evil Fish)
:lol: feeling stupid now...

Don't understand.
Don't know what else to suggest.

The simplification I suggested at the end of my previous post is likely to cover everything at A-level, but it's not foolproof.
0
reply
L'Evil Fish
Badges: 18
Rep:
?
#13
Report 7 years ago
#13
(Original post by ghostwalker)
Don't know what else to suggest.

The simplification I suggested at the end of my previous post is likely to cover everything at A-level, but it's not foolproof.
Ha don't worry, I get what it means but things like a cap and that. Can't see the latex
0
reply
ghostwalker
  • Study Helper
Badges: 17
#14
Report 7 years ago
#14
(Original post by L'Evil Fish)
Can't see the latex
:holmes: Curious.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (53)
17.67%
I have experienced some financial difficulties (83)
27.67%
I haven't experienced any financial difficulties and things have stayed the same (115)
38.33%
I have had better financial opportunities as a result of the pandemic (39)
13%
I've had another experience (let us know in the thread!) (10)
3.33%

Watched Threads

View All
Latest
My Feed