Question

what is the domain and range of both these equations

1) 3x / 5-|7-4x|

2) square root of -1 / (3-x)(2-x)

Answer 1

The domain is all of the values which can be put into the equation. For 1) can you see which values of x we would not be allowed to input?

Answer 2

@Traxter my answer is
1. domain: [7/4, ~)
range : (~ ,5]
not so sure of it's right

then the range of the second one, i don't know.

Answer 3

For the first one, we can't have the denominator equal to zero. So solve 5-|7-4x|=0 to find which values of x will be excluded from the domain. See how you do with that.

Answer 4

Domain= R-{1/2}

Answer 5

since it is in p/q form denominator should not be equal to 0

Answer 6

@mukushla

Answer 7

I think you missed a value of x in the first question which would make the denominator 0.
\[5-|7-4x|=0\] implies \[|7-4x|=5\]
So \[7-4x=5\] and \[4x-7=5\]
Hence the domain is R - {1/2,3}

Answer 8

Oh....Yes...there are two cases...

Answer 9

Indeed :) Now are you able to use the domain we have found to find the range?

Answer 10

i forgot..that

Answer 11

for range put y =3x / 5-|7-4x|

Answer 12

@satellite73

Answer 13

@Traxter ...wat abt Range..

Answer 14

5y - 7y + 4xy = 3x -------------Case I

Answer 15

Rearrange the equation so that you get x in terms of y (you will get 2 of them because of the modulus) and find the domain of y.

Answer 16

-2y = 3x - 4xy

-2y = x(3-4y)

x = -2y/(3-4y)

3- 4y not equal 0

y not equal 3/4 -------- 1 st soln

Answer 17

Case II

5y + 7y - 4xy = 3x

12y = x(3 + 4y)

x = 12y/(3+4y)


3+ 4y not equal 0

y not equal -3/4

Answer 18

@Traxter .... Do...u think mine is correct

Answer 19

Yes :) Good work.

Answer 20

so what's the range? the union of both of them ? @Traxter ?

Answer 21

Yes, the range will be R - {the values of y we found we could not have}

Answer 22

and how about for the second one ? @Traxter @Yahoo!

is it going to be

domain: (~,3)U(3,2)U(2,~) ?
range: [0,~) ?

Answer 23

For the second one we cannot have the denominator equal to zero, that is, we cannot have (3-x)(2-x)=0. So x cannot be 3 and x cannot be 2. Also, since we have the square root of negatives we will be in the complex plane. So the domain will be C-{2,3}

Answer 24

yeah, and the range ?

Answer 25

Have a go at working it out yourself, using the same method we used before. Find x in terms of y (or x in terms of f(x), whatever notation you're using) and see which values of y we can't have.

Answer 26

i'm not sure. well i'm sure i did it wrong.

Answer 27

@Traxter