find the domain and range? 10 points....?

for # 1:

if you meant a) y = (5/2x)-3:

domain: all x except x=0

range: all real numbers

if you meant b) y = 5/(2x-3)

domain: 2x - 3 /= 0

all x except x = 3/2

range: all real numbers

2. y = sqrt [x(x-2)]

domain:

x(x-2) >= 0

a. x >=0 and x-2 >= 0

x >=0 and x >=2

b. x<0 and x-2 < 0

domain is all x < 0 OR all x >=2

range: all real numbers

Even before considering the domain and range of each function, it would be helpful to remove any ambiguity in your expressions.

1.)y= 5/2x-3 The way you've typed it can be taken to be

a) 5/(2x-3)

b) 5/(2x) - 3 or possibly

c) (5/2)x - 3 or maybe even

d) (5/2)(x - 3) and the likelihood is that each interpretation produces a different answer.

Similarly

2.)y=the square root of x^2-2x could be

a) the square root of [x^2-2x] or maybe

b) (the square root of [x^2]) - 2x

So please help us help by judiciously using grouping symbols.

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I feel compelled to add (because other answers are misleading) that if 2) is a) the square root of [x^2-2x] then your domain is x<=0 or x>=2.

An alternative notation for your domain is

(-infinity, 0] U [2, +infinity)

Notice that (0,2) or 0<x<2 is NOT part of your domain.

Your range for 2) would be [0, +infinity) or 0 < = x

y=5/(2x-3)

=> 2x-3 is non zero

i.e. x is not equal to 3/2

so x>3/2 or x<3/2

& y is real

sqrt(x^2-2x)

x(x-2)>=0

x>=0, x>=2 or x<=0,x<=2

i.e. x>=2 or x<=0

& y is a +ve real

1.)

x = all real but x cannot = 3/2

y = all real but y cannot = 0

2.)

cant be bothered thinking atm... ill post later...