Question

Find the domain of the function
f(x)=x-5/2x-1

Answer 1

\(f(x)=\frac{x-5}{2x-1}\)
?

Answer 2

the domain is all real numbers

Answer 3

no its not

Answer 4

yes it is

Answer 5

and dont just give answers

Answer 6

no its not, please stop....

Answer 7

ok

Answer 8

@heatherjoy

The only problem with rational functions composed of polynomials is that the denominator CANT equal to \(0\). SO in this situation \(2x-1\ne0\)
what does this tell us that x CANT be?

Answer 9

just giving me the answer doesn't help. i don't know how you got it and i have to show my work

Answer 10

do you understand what I wrote?

Answer 11

nope

Answer 12

so you have \(\frac{x-5}{2x-1}\) correct

Answer 13

?

Answer 14

i can't see the top part of your post

Answer 15

im confused..

Answer 16

there is no top part, I just put space there

Answer 17

is that a 5? it looks like a b

Answer 18

\(\huge \frac{x-5}{2x-1}\)

Answer 19

ok, yes

Answer 20

so the only problem here is that we cant divide by \(0\)

Answer 21

right

Answer 22

in other words \(\huge 2x-1\ne 0\)

Answer 23

ok

Answer 24

now treat that \(\ne \) like it was an \(=\) and solve for x

\(2x-1\ne 0 \implies 2x\ne 1 \implies x\ne \frac{1}{2} \)

Answer 25

so our domain is \(\{x\ |\ x\ne\frac{1}{2}\}\)

Answer 26

or in words

All \(x\) such that \(x\) is NOT \(\frac{1}{2}\)

Answer 27

what is that fraction? i'm sorry. i can't see it

Answer 28

\(\huge \frac{1}{2}\)

Answer 29

\(\huge 2x-1\ne 0 \implies 2x\ne 1 \implies x\ne \frac{1}{2}\)

Answer 30

ok

Answer 31

can you tell me wht the steps are then?

Answer 32

this is showing me what not to do

Answer 33

?

Answer 34

i don't understand

Answer 35

those are all the steps


this is how I would answer this question:

\(\huge 2x-1\ne 0 \implies 2x\ne 1 \implies x\ne \frac{1}{2}\)

so our domain is \(D=\{x \ | \ x \ne \frac{1}{2}\}\)

Answer 36

oh. wow. i'm really confused

Answer 37

asking for the domain is the same thing as asking

\(\huge\text{what is x allowed to be}?\)

Answer 38

an integer

Answer 39

nono

Answer 40

read the whole thing

Answer 41

when you ask for the domain of a function you are asking what is x allowed to be, then we notice that x can be anything we want as long as we dont divide by 0. The way you say that "mathematically" is

\(\huge 2x-1\ne 0\)

Answer 42

ive never seen this before. it looks like it's telling me x isn't allowed to be those things. i've never seen that symbol

Answer 43

ok you dont have to write it like that, you can just say all x except 1/2

Answer 44

i'm new to algebra, sorry

Answer 45

its no problem

Answer 46

most of the time we have to list domain in terms in what it is not allowed to be, because we cant list out every real number (there are infinite of them)

Answer 47

ohhhhh

Answer 48

so instead of saying what it can be, we say it can be anything except some certain numbers

Answer 49

ok that makes sense

Answer 50

so we decided that the denominator CANT be 0

our denominator is \(2x-1\)

Answer 51

so the answer is what you wrote, simply because x isn't allowed to be anything else in the equation?

Answer 52

our denominator CANT be 0

\(\downarrow\)

\(\huge 2x-1\ne 0\)

Answer 53

ok, so this is the beginning of solving?

Answer 54

if we solve this we get that \(x\) cant be 1/2

Answer 55

truthfully, i don't knw what a domain is, or what the problem is even asking me to do

Answer 56

1/2 is the ONLY thing that will make the denominator 0. thus our domain is all real numbers EXCEPT 1/2

Answer 57

ok

Answer 58

domain is just set of numbers we are allowed to plug into the function and get an answer that makes sense

the domain of \(f(x)=\frac{1}{x}\) is all numbers EXCEPT 0

the domain of \(f(x) = 3x + 4\) is all real numbers, because there is no chance to divide by 0

the domain of \(f(x) = \frac{x+3}{x-7}\) is all real number EXCEPT 7, because x=7 makes the denominator be 0

Answer 59

ok, so my answer is just that then?
in this instance

Answer 60

what about the 2?

Answer 61

so we are only concerned with when 2x-1 is equal to 0

so we see when that happens

\(\huge 2x-1=0 \implies 2x=1\implies x=\frac{1}{2}\)

Answer 62

so is this a first step? or the only step in this problem?

Answer 63

are you still there?

Answer 64

are you still there?

Answer 65

look up to where I said this is how I would answer the problem.