Question

find the domain and range g(z)= 1 over square root 4-z square

Answer 1

find the value of z that will make the equation undefined...
sq root (4-z^2) = 0
4 - z^2 = 0
4 = z^2
z = +/- 2

so...your domain is all values of x except for 2 and -2...which can be written as...(-inf, -2)U(2, inf)

as for the range...cross multiply.
y(sq root 4-z^2) = 1
4 - z^2 = 1/y^2
z^2 = 4 - 1/y^2

z^2 = (4y^2 - 1 )/y^2
get the sq root

z = sq root (2y +1)(2y-1)/y

anyways...your range is everything below and equal to zero...so your range is

(-inf, 0)U(0, inf)

Answer 2

how you throw away the square root to find range?

Answer 3

y(sq root 4-z^2) = 1

divide both sides by y

sq root 4-z^2 = 1/y

square both sides...
4-z^2 = 1/y^2

Answer 4

hmmm
\[\frac{1}{\sqrt{4-z^2}}\] need
\[4-z^2>0\]
\[(2-z)(2+z)>0\] i think this is
\[(-2,2)\] as the domain

Answer 5

it is a square root right?

Answer 6

is z a real variable?

Answer 7

ooooooooh range i see
ok that would be
\[(\frac{1}{2},\infty)\]

Answer 8

reason as follows.
this beast can get as large as possible by making the denominator small, that is make x close to 2 or -2

the denominator is largest when x is 0, and when x is 0 you get
\[\frac{1}{\sqrt{4}}=\frac{1}{2}\] so that is the minimum value of the function

Answer 9

so to be more precise, range is
\[[\frac{1}{2},\infty)\] or
\[\frac{1}{2}\leq y<\infty\]

Answer 10

actly which answer the correct one??? im blur

Answer 11

\[g(z)=\frac{1}{\sqrt{4-z^2}}\] Domain
\[(-2,2)\]
Range
\[[\frac{1}{2},\infty)\]

Answer 12

r u sure????

Answer 13

why u say that x>0