range q (1 Viewer)

[Green Yoda]

for y=1/sqrt(2x+9) find domain and range

I got the domain as x>-4.5
but how do I find the range?

 

[kawaiipotato]

 

[Green Yoda]

answer is y>0
I dont know how to get it without graphing it

 

[leehuan]

answer is y>0
I dont know how to get it without graphing it

Use what kawaiipotato said and the simple fact that

1/sqrt(ax+b) = y where a=2 and b=9

y>0

You should be well aware of the first line in his working, that is, anything under a square root is positive or equals to 0.

 

[Green Yoda]

ahh I think i got it..y cannot be negative as there is a square root there so its always positive and it cant be 0 because to have a 0 in the denominator it will be undefined so y>0

 

[Drongoski]

ahh I think i got it..y cannot be negative as there is a square root there so its always positive and it cant be 0 because to have a 0 in the denominator it will be undefined so y>0

No no no!

It cannot be zero because the numerator ('1') can never be zero. When you have an expression: numerator/denominator, this expression becomes 0 only when the numerator becomes 0; nothing to do with the denominator, except you must ensure it is not 0.

 

[Ambility]

ahh I think i got it..y cannot be negative as there is a square root there so its always positive and it cant be 0 because to have a 0 in the denominator it will be undefined so y>0

You are thinking about the function's domain and what x values it cant have.
We need to be thinking about what y values it will never be.



The denominator is a principal root, which is always positive. One on something that is always positive is going to be always positive. Also, as Drongoski pointed out, no matter what x value you put into it, it is impossible to make the function equal to zero due to the one on top. The range is therefore:

 

[Green Yoda]

You are thinking about the function's domain and what x values it cant have.
We need to be thinking about what y values it will never be.



The denominator is a principal root, which is always positive. One on something that is always positive is going to be always positive. Also, as Drongoski pointed out, no matter what x value you put into it, it is impossible to make the function equal to zero due to the one on top. The range is therefore:

ahh I clearly understand now, thank you

 

[InteGrand]

Everyone here has explained that y needs to be positive, but to truly claim the range is y>0, we would also need to show that is indeed every positive y value can be attained (e.g. for the function y = 1/sqrt(1+x^2), we also have y > 0 always, but the range is not just y > 0, because in fact y can't ever exceed 1).

To show that for any y > 0, there exists an x in the domain such that f(x):= 1/sqrt(2x+9) = y, note that we can take x = ((1/y)^2 – 9)/2 (this is clearly well-defined and in the natural domain for any positive y, where the natural domain is {x: x > -9/2}). Then f(x) = y, by inspection or substitution. Hence the range is indeed {y: y>0}.