[AQA] C3: Range of a function

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.

(edited 10 years ago)

Reply 2

Dental Plan

OP

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.

Reply 3

WhiteGroupMaths

9

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.

Reply 4

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.

Reply 5

Can I ask....

When combining functions, what are the rules with regards to domains and ranges?

Reply 6

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,

(edited 10 years ago)

Reply 7

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?

Reply 8

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.

(edited 10 years ago)

Reply 9

maggiehodgson

14

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.

Reply 10

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.

feeling stupid now...

Don't understand.

Reply 11

Original post by L'Evil Fish

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.

Reply 12