how to find the domain and range C3

how, i find it so hard and to lose 4 marks for both is alot.

You know what the domain and range of a function are, right? It would help if you gave an example question, and any working you've done.

Reply 2

ajayhp

1/(x+1)

i thought it would be x>-1

(edited 13 years ago)

Reply 3

StephenP91

Original post by ajayhp

1/(x+1)

i thought it would be x>-1

Haven't they given a set interval for the domain as x can be any value from inf to -inf (except -1 of course)

(edited 13 years ago)

Reply 4

ajayhp

Original post by StephenP91

The domain itself it all negative and positive numbers, bar -1.

Why? how would you do it?

Reply 5

Dragon

You can't just "find" the domain of a function. The domain could be pretty much anything. The domain of your example function could be

\{1\}

\mathbb{N}

\{1,5,3423,134\}

. Basically any set of numbers not including

-1

. The sensible or natural domain for this function is

\mathbb{R} \backslash \{-1\}

, which is almost certainly the domain you want.

A common question is: Given some function with some domain, what is its range?

Now you gave

f(x) = \frac{1}{x+1}

as an example. You didn't specify a domain. Let's say the domain is

D = \{0,1,2\}

. Then the range is the image of the domain, or

\{f(x) | x \in D\} = \{\frac{1}{3},\frac{1}{2},1\}

(edited 13 years ago)

Reply 6

ElMoro

Original post by Dragon

You can't just "find" the domain of a function. The domain could be pretty much anything. The domain of your example function could be

\{1\}

\mathbb{N}

\{1,5,3423,134\}

. Basically any set of numbers not including

-1

. The sensible or natural domain for this function is

\mathbb{R} \backslash \{-1\}

, which is almost certainly the domain you want.

A common question is: Given some function with some domain, what is its range?

Now you gave

f(x) = \frac{1}{x+1}

as an example. You didn't specify a domain. Let's say the domain is

D = \{0,1,2\}

. Then the range is the image of the domain, or

\{f(x) | x \in D\} = \{\frac{1}{3},\frac{1}{2},1\}

I think you're confusing the OP with notation :wink2:

Reply 7

ajayhp

Original post by Dragon

You can't just "find" the domain of a function. The domain could be pretty much anything. The domain of your example function could be

. Basically any set of numbers not including

. The sensible or natural domain for this function is either

, one of which is almost certainly the domain you are looking for.

A common question is: Given some function with some domain, what is its range?

Now you gave

as an example. You didn't specify a domain. Let's say the domain is

. Then the range is the image of the domain, or

domain x>3
how would you find the range

Reply 8

ElMoro

Original post by ajayhp

domain x>3
how would you find the range

Ok, you can see as x gets bigger, f(x) is getting smaller. Right?

So the maximum value would be when x=3, f(x)=1/4
But since x never actually equals 3, f(x) never actually equals 1/4
So the range is: f(x)<1/4

Hope this made sense :biggrin:

Reply 9

ajayhp

would x=3 be the minimum as x has to be greater than 3

Reply 10

ElMoro

Original post by ajayhp

would x=3 be the minimum as x has to be greater than 3

Did you understand any of my post or do you want me to start again? :smile:

Reply 11

ajayhp

Original post by ElMoro

Did you understand any of my post or do you want me to start again? :smile:

sorry about this, i know x cannot equal 1, but why do we put 3 in since it says that it should be greater than 3. this is waht i dont understand, and also what would happen if we had an interval for the domain like 2<x<5

Reply 12

beelz

wouldnt the range be 0<f(x)<0.25 if the domain is x>3?

Reply 13

ElMoro

Original post by ajayhp

Ok, I will explain in more datail soon. :smile:

Original post by beelz

wouldnt the range be 0<f(x)<0.25 if the domain is >3?

Oh crap, I forgot that bit. :shock:

Yeah, that's right. :biggrin:

Reply 14

Dragon

Original post by ajayhp

As ElMoro said, the function decreases as x increases. So we know that if

f(3) = \frac{1}{4}

, then the range will be all real numbers strictly less than

\frac{1}{4}

(and greater than 0).

Note the strict inequality.

f(x)

can never equal

\frac{1}{4}

because the domain is

x>3

(strictly greater than). However the function can take a value extremely close to, but smaller than,

\frac{1}{4}

if x is just a tiny bit smaller than 3.

Another way of thinking about it is that

\frac{1}{4}

is the upper bound of values the function can take, or the upper bound of the range. You need to evaluate f(3) in order to find this bound. Similarly, for the domain

2<x<5

, you will need to evaluate f(2) and f(5) to find upper and lower bounds for the range.

(edited 13 years ago)

Reply 15

nuodai

Let's say you have some function

f : x \mapsto f(x)

and you're told to find the natural domain and range. The range depends upon the domain, so we'll sorry about that later and concentrate on the domain for now.

The natural domain is the set of real numbers for which the function "makes sense". So you always start with

\mathbb{R}

, and then take away anything where the function doesn't make sense. Graphs help for doing this; if you draw a graph of

y=f(x)

, then the places where the function doesn't make sense will be the places where the graph doesn't go over or under that point.

So for example if

f(x) = \frac{1}{x}

then the point

x=0

doesn't make sense, so the natural domain is

\mathbb{R} \smallsetminus \{ 0 \}

, and this is reflected in the graph, since there is no point on the graph directly above or below x=0. If

f(x) = \sqrt{x-2}

, then because we can't take the square root of a negative number, the domain will be the set of points where

x-2 \ge 0

, i.e.

x \ge 2

. This is also reflected in the graph of the function, since you can only draw it for

x \ge 2

; there is no point on the graph directly above or below any value of

x

when

x<2

.

If you're given part of the domain, then you need to do exactly the same thing, except that instead of starting with

\mathbb{R}

you start with what you've been given. So, if like here you're given that

x>3

, then you just forget that any values of

x \le 3

exist; we only look at the points greater than 3 which make sense. So if we're given

f(x) = \dfrac{1}{x}

and given that

x>3

, then we're fine; the only point where it doesn't make sense is x=0 and 0<3, so we don't have to worry about it... so in this case, the domain is simply

x>3

.

But say we're given

\sqrt{x-5}

and told to find the natural domain for

x \ge 3

. If we just had

\mathbb{R}

then we'd just say that we need

x - 5 \ge 0

, so the domain is

x \ge 5

. Just because we've changed the set of values that x can take doesn't mean the function is any different, and since

5 \ge 3

this is something we have to worry about, so the domain is still

x \ge 5

. Contrast this with

\sqrt{x-2}

which, if it's specified that

x \ge 3

, has natural domain

x \ge 3

, rather than

x \ge 2

as it was in the case when we look at the whole of

\mathbb{R}

.

Next up is the range. The range is just the set of values that we can make the function be equal to by plugging in values of the domain. To find this, we can look at the graph again, and the range is the set of y-values for which there is some point on the graph horizontally across from it. So if

f(x)=x^2

then the range is

f(x) \ge 0

(there are no points horizontally across from any negative numbers), if

f(x)=\sin x

then the range is

-1 \le f(x) \le 1

, and if

f(x)=x^3

then the range is

f(x) \in \mathbb{R}

.

Finding the range is a bit more difficult when we have some specification on the domain. For example, consider

f(x) = \sqrt{x-2}

when we're told that

x \ge 3

. In this case, the natural domain (subject to the requirement

x \ge 3

) is just

x \ge 3

. If we didn't have this specification, the range would simply be

f(x) \ge 0

and the domain would be

x \ge 2

, so in this case it makes sense to work with that and then "chop away" the bits that we're not allowed to have. We're not allowed to have any bits where

2 \le x < 3

. But notice that

\sqrt{x-2}

is an increasing function, so the interval

2 \le x < 3

corresponds exactly to the interval

\sqrt{2-2} \le f(x) < \sqrt{3-1}

, i.e.

0 \le f(x) < 1

. So, our resulting range is

f(x) \ge 0

with

0 \le f(x) < 1

chopped off... that is, the range is simply

f(x) \ge 1

.

Hope this helps... I understand it might be a bit of a bitch to read through.

[This question comes up quite a lot and for some reason I always type out a new response, so for future reference I'm going to tag this post with the word SPAGHETTI so that I can find it easily in the search box and copy+paste it next time this comes up.]

(edited 13 years ago)