#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..
#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
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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
#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
Quote by Kelvster xp
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
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#9
thanks a lot!!! rbmason!! your great!!! but wait one more question why is it that the restrictions are 3-negative numbers?
#10
Quote by Kelvster xp
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.
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#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.
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#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)
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