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

1. anonymous

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)

2. anonymous

how you throw away the square root to find range?

3. anonymous

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

4. anonymous

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

5. anonymous

it is a square root right?

6. anonymous

is z a real variable?

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

actly which answer the correct one??? im blur

11. anonymous

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

12. anonymous

r u sure????

13. anonymous

why u say that x>0