Question

how am i suppose to determine the domain and range for the function: y= the square root of (x-4)? i dont want the answer i just want to understand the problem

Answer 1

domain is all usable values for "x"

the range is all the results for "y" that you get by using the elements of the domain

Answer 2

ask yourself, does the square root function have any restrictions? what makes it go bad?

Answer 3

wait so the domain is all real numbers? why?

Answer 4

if the range and domain of the inverse is easier to find; that would be another method

Answer 5

does the square root function have an restrictions? for example

sqrt(-4) = ??
sqrt(0) = ??
sqrt(9) = ??

can we solve these?

Answer 6

no

Answer 7

we cant solve any of them?

Answer 8

well isnt the sqr of -4 an imaginary number?

Answer 9

or something?

Answer 10

yes it is, if we are allowed to step out of the reals

Answer 11

The question is not very well worded...:|
It's supposed to say that we have a "Real function".

Answer 12

if your allowed to use the set of Complex numbers, then the domain is all complex values

Answer 13

@ParthKohli yeah thats right

Answer 14

or rather, if y is allowed to be an imaginary number; then the domain can include the set of reals

Answer 15

\[\sqrt{u(x)}~:~u(x)=x-4\]
\[u(x)\ge0\]
\[x-4\ge0\]
will define our domain if we assume sqrt to be a real valued function

Answer 16

wait im confused

Answer 17

youll have to be more specific about your confusion

Answer 18

i want to know something concrete about how im suppose to solve problems like this kind of like a rule or certain steps because i dont understand the concept

Answer 19

domain is the set of elements that are allowable:

lets say that you have 20 dollars to spend in the store; your domain is then $20

the range of options for you to purchase is then within the values of free to $20 less taxes

Answer 20

oh ok

Answer 21

the steps for the mathical stuff is:

determine usable values for the given function

determine the values of "x" that give us those values

if the domain has a least value, plug that in to the least value of the range
if the domain has a highest value, plug that in to the highest value of the range

Answer 22

so, what values, if any, do we have to omit from the square root function for it to be give us real values?

Answer 23

isnt there alot of options to plug in?

Answer 24

i spose we could name them one by one, or it might be easier to group them into a specific set that defines what they all have in common

Answer 25

oh so like all real numbers? does that go for the range too?

Answer 26

all real numbers is not correct

Answer 27

sqrt(-4) does not produce a real number does it?

Answer 28

ok will it be all real numbers greater than -4?

Answer 29

we might be thinking of 2 different things here, im not focused on the (x-4) part just yet

im just trying to define the values that either make sqrt a real function, or not

Answer 30

sqrt(-3) is not a real value; sqrt(-2) is not a real value; sqrt of any negative number does not produce a real value

Answer 31

negative number are less than 0

so, when is x-4 < 0 ?

or to say the same thing differently

when is x-4 >= 0 ?

Answer 32

teh second setup will be more useful to us to define the domain with since it produces all the values of x that we can actually use

Answer 33

do you mean greater than or equal to zero for the second setup ? so i guess 4 is the lowest number that can be plug into the second setup for it to equal 0

Answer 34

yes

Answer 35

so all numbers greater than or equal to four?

Answer 36

that is our domain, yes

Answer 37

the range is found by using the domain

at x=4; y=0
at x=5; y=1
at x=8, y=2
at x=13; y=3

without getting into calculus, its hard to determine if there is a maximum value for y; but i already know that \(\sqrt{\infty-4}=\infty\)

Answer 38

\[y=\sqrt{x-4}\]
\[y'=\frac{1}{2\sqrt{x-4}}\]
since y' can never be zero in this case, there is no maximum, and the function is always increaseing

Answer 39

gotta get to class, good luck ;)

Answer 40

ok thanks for all your time and help