#1

so we had our assignment today and i have only few ideas on how to answer this sh*t..

FInd the range and domain of the following functions:

1. y=x

2. 1/x

3. y=2x^2+4x-1

4. y= root of x - 3

now i know that the domain in number four is the set off all x such that s is greater than or equal to zero.. since x should not be a negative number... now.. how do i find the range?? and the domain of the numbers 1-4? i really need help..please.. somebody..

FInd the range and domain of the following functions:

1. y=x

2. 1/x

3. y=2x^2+4x-1

4. y= root of x - 3

now i know that the domain in number four is the set off all x such that s is greater than or equal to zero.. since x should not be a negative number... now.. how do i find the range?? and the domain of the numbers 1-4? i really need help..please.. somebody..

#2

grade 10 functions. oh boy, bad memories

#3

i know.. do you know the answer??? or an explanation that my tiny little brain could understand?

#4

1:

d = all reals

r = all reals

2:

d = all reals such that x != 0

r = all reals except 0

3:

d = find the factors (x + ?)(x - ?) example. find x = +- ?

so d = all reals such that x != ?

r = usually > 0 may be different

4:

d = all reals such that x > 3

r = all reals such that y > 0

edit:

my 3 is wrong,

d = all reals

r = y > number (have to solve the equation for y being min

d = all reals

r = all reals

2:

d = all reals such that x != 0

r = all reals except 0

3:

d = find the factors (x + ?)(x - ?) example. find x = +- ?

so d = all reals such that x != ?

r = usually > 0 may be different

4:

d = all reals such that x > 3

r = all reals such that y > 0

edit:

my 3 is wrong,

d = all reals

r = y > number (have to solve the equation for y being min

*Last edited by rbmason at Jun 17, 2008,*

#5

1. range & domain = all reals

2. range = all reals. domain = {-infinity<x>0}U{0<x>infinity}

3. range = > -3. domain = all reals

4. if you meant (x-3)^(1/2) then the range is >0 domain is > 3

2. range = all reals. domain = {-infinity<x>0}U{0<x>infinity}

3. range = > -3. domain = all reals

4. if you meant (x-3)^(1/2) then the range is >0 domain is > 3

#6

er in number four wouldnt it be the D= all reals such that x is greater than or equal to zero? or my mind just kinda mess up..?

#7

why is it that the range in number 4 is >0 and the domain is >3????? i dont get it..?? help again!

#8

er in number four wouldnt it be the D= all reals such that x is greater than or equal to zero? or my mind just kinda mess up..?

well, in real world math, not reals math, it would be x = anything,

but since you can't have imaginary numbers,

any x < 3 would result in jK, where K is some constant.

j is imaginary = sqrt(-1)

edit

the range is > 0 because you can't plot a negative sqrt on the xy plane.

that is, for that function, y will never be less than 0

#9

thanks a lot!!! rbmason!! your great!!! but wait one more question why is it that the restrictions are 3-negative numbers?

#10

thanks a lot!!! rbmason!! your great!!! but wait one more question why is it that the restrictions are 3-negative numbers?

which one are you talking about?

I don't understand your question.

#11

i mean in number 4.. i still dont get why x is greater than 3...

#12

ok, if x is any number, say, greater than 3, like 6

sqrt (5 - 3) = sqrt (2), which although irrational, is real (not imaginary)

if x is equal to 3, sqrt (3 - 3) = sqrt (0) = 0

if x is less than 3, say, 0

y = sqrt (0 - 3) = sqrt (-3)

problem : can you have the sqrt of a negative number?

for your class, no. it's not possible.

in fact it is, but it uses an imaginary number, j , to represent it.

i.e.: sqrt(-9) = j3

so for this problem, x has to be greater or equal than 3 to stay a real number, not necessarily rational, just has to be real.

so I was wrong, the answer is in fact

x >= 3 for y to be real

which is the definition of the domain.

sqrt (5 - 3) = sqrt (2), which although irrational, is real (not imaginary)

if x is equal to 3, sqrt (3 - 3) = sqrt (0) = 0

if x is less than 3, say, 0

y = sqrt (0 - 3) = sqrt (-3)

problem : can you have the sqrt of a negative number?

for your class, no. it's not possible.

in fact it is, but it uses an imaginary number, j , to represent it.

i.e.: sqrt(-9) = j3

so for this problem, x has to be greater or equal than 3 to stay a real number, not necessarily rational, just has to be real.

so I was wrong, the answer is in fact

x >= 3 for y to be real

which is the definition of the domain.

#13

OOPs! im very sorry my fault its suppose to be the (root of x)-3... im terribly sorry.. but im still very thankful that you helped me understand our assingment,... thanks.. so therefore it would be x =>0?

#14

yeah, that's it, now you'veg ot it.

but the domain is > 0

and the range is -3 (because x can't be less than zero, the lowest value of y is -3)

but the domain is > 0

and the range is -3 (because x can't be less than zero, the lowest value of y is -3)

#15

thanks again Rbmason.. you really helped a lot.. cheers!