FP1 help: i seriously don't understand ranges and domains :( will rep?

i am fine rearranging functions etc but when it comes to finding the range or domain i have no idea what that means or how to find it and the internet isnt helping :frown:

the function f(x) = (2x +3)/(x-1)

x something to do with real values (funny e letter then the weird R), and x >1

a) find f^-1 (x) i have done this, and found y = (x+3)/(x-2)
what else do i need to write after it? ill write the real values thing, but what else

b) find the range of f^-1 (x)
c) find the domain of f^-1(x)

seriously don't have a bloody clue :frown:

help really appreciated because i just dont understand at all

cheers,

jb

Reply 1

fps_cam

the range is the values of x you can put in to get a numerical result

the domain is these results

eg above, x cannot equal 1 as you would be dividing by zero, which is a no-go

Reply 2

lee_91

10

domain is the values u can put into the function for it to work

range is the values u can get out when said domain is applied

draw a graph, and its the values for which the function is between in x and y respectively

Reply 3

sunnyhocl

range is the range of x-values in which the function exists.
domain is the range of y-values in which the function exists.
for example the fucntion y = x^2 has a domain of x greater than or equal to zero because there isn't anything that belongs to this curve under the x-axis.
and the range of the function y = sin^-1 x has a range of -1<x<1 because there is nothing beyond this set of numbers that belongs to this function.
hope that helps

Reply 4

mathperson

20

jumblebumble

i am fine rearranging functions etc but when it comes to finding the range or domain i have no idea what that means or how to find it and the internet isnt helping :frown:

the function f(x) = (2x +3)/(x-1)

x something to do with real values (funny e letter then the weird R), and x >1

a) find f^-1 (x) i have done this, and found y = (x+3)/(x-2)
what else do i need to write after it? ill write the real values thing, but what else

b) find the range of f^-1 (x)
c) find the domain of f^-1(x)

seriously don't have a bloody clue :frown:

help really appreciated because i just dont understand at all

cheers,

jb

The range means what y-values is the function between and the domain means what x values is the function between.

Example:

f(x) = sin(x) has range
-1 (greater than or equal to) y (less than or equal to) 1
and has domain
-infinity (greater than) x (less than ) infinity.

xER means that x can take any real value (basically any number : 1, 2, -1, 1/2, pi, sqrt2, ....)

a) I don't think you need to write anything else. I havn't checked the answer though.

c) The range is xER (remember what I said this means) x=/= 2 (x does not equal 2). This means that x can take any real number apart from 2 (because if x=2 it means you divide by infinity which is undefined).

b)I think the range is -infinity (greater than) y (less than) infinity. I havn't drawn the graph though.

When doing this sort of thing, look at the denominator and see what that tell you, like I did in part c) of the question, but also draw a sketch of the function on a set of axes if you can because it helps alot.

good luck.

EDIT - I just realised I made a small mistake, the question says xER, x>1, so when you draw your diagram in order to find the domain and range of f^-1(x), do take this into account.

Reply 5

AnonyMatt

15

The funny E means "is an element of" or something. And the big R means the set of real numbers.
So x is any real number, greater than 1.

You don't need to write anything after your f^-1(x).

Note that the range of f^-1(x) is the domain of f(x), and vice versa.

So either find the range of f(x) for c), or determine the range of f^-1(x).

The range is simply all the y values that a function can take. I assume you've done work on horizontal asymptotes?

Make x = infinity. What does the function tend to?

Sketch the graph. It should be clear what the range is then.

Reply 6

plonplon

to put it bluntly domain is what you put in range is what you get out

Reply 7

mathperson

20

AnonyMatt

The funny E means "is an element of" or something. And the big R means the set of real numbers.
So x is any real number, greater than 1.

You don't need to write anything after your f^-1(x).

Note that the range of f^-1(x) is the domain of f(x), and vice versa.

So either find the range of f(x) for c), or determine the range of f^-1(x).

The range is simply all the y values that a function can take. I assume you've done work on horizontal asymptotes?

Make x = infinity. What does the function tend to?

Sketch the graph. It should be clear what the range is then.

you don't 'make x=infinity', you determine the limits (what the function tends towards) as x tends towards both + and - infinity, this gives the range.

Reply 8

DerBoy

7

jumblebumble

i am fine rearranging functions etc but when it comes to finding the range or domain i have no idea what that means or how to find it and the internet isnt helping :frown:

the function f(x) = (2x +3)/(x-1)

x something to do with real values (funny e letter then the weird R), and x >1

a) find f^-1 (x) i have done this, and found y = (x+3)/(x-2)
what else do i need to write after it? ill write the real values thing, but what else

b) find the range of f^-1 (x)
c) find the domain of f^-1(x)

seriously don't have a bloody clue :frown:

help really appreciated because i just dont understand at all

cheers,

jb

Don't rep me, I'd rather you didn't.

f^{-1}(x)

is the inverse of the function f(x).
Consider the function f(x)=3x where x is an integer between 0 and 5 inclusive.
The domain of f(x) would be all integers (Z) from 0 to 5, i.e. {0,1,2,3,4,5}.
The range of f(x) would be the set of the triples of all elements in the domain, and nothing else; i.e. {0,3,6,9,12,15}.

The function f(x) as defined picks the element from the domain and outputs the corresponding element in the range. For example f(3)=9.

The inverse of that function, written

f^{-1}(x)

, does the opposite of f(x). It picks an element of the range and goes back to the corresponding element in the domain. For example

f^{-1}(15)=5

.

If f(x)=x+3, then

f^{-1}(x)=x-3

.
If f(x)=5x, then

f^{-1}(x)=\frac{x}{5}

.
If

f(x)=x^2

, then

f^{-1}(x)=\sqrt x

If f(x)=5x+3, then

f^{-1}(x)=\frac{x-3}{5}

(You are doing the opposite of the original function- if the original functions says add, the inverse says subtract. If the original says multiply, the inverse says divide. If the original says square, the inverse says square root. If the original says multiply by 5 then add 2, then the inverse says subtract 2 then divide by 5.)

You may be confused about the last (inverse) function above.
An inverse function does the opposite of what the original function does, but it does it in a reverse order. This didn't matter for when the original function only did one thing.
Written in mathematical terms:
If

f(x)=a_1(a_2(....(a_{n-1}(a_n(x)))....)))

, then

f^{-1}(x)=a_n(a_{n-1}(....(a_2(a_1(x)))....)))

Reply 9

AnonyMatt

15

mathperson

you don't 'make x=infinity', you determine the limits (what the function tends towards) as x tends towards both + and - infinity, this gives the range.

Excuse my wording, but it's how it's often taught at A level...

And x > 1 so it cannot tend towards -infinity.

Reply 10

mathperson

20

AnonyMatt

Excuse my wording, but it's how it's often taught at A level...

And x > 1 so it cannot tend towards -infinity.

ok.

What you have to do (more precisely) is see what the function tends towards when you allow x to tend towards the upper and lower 'bounds' in this case 1 and infinity.

FP1 help: i seriously don't understand ranges and domains :( will rep?

Related discussions

Latest

Trending

Trending

Articles for you