How do you find the domain and range of 1/square root of x-9
I think that this is what you intended?
f(x)=1√x−9
Let see..
you cannot divide by 0 and the square root of a real number cannot be neg so the range must be greater than 0
x-9 cannot be negative so x>9
so the range is f(x)>0
and the domain is x>9
Here is the funtion
How do you find the range of 1/square root of x-9
Hello M!
1√x−9=1√1x9=√x9=x4√x
sorry!
so
1√x−9=√x−9x−9
More is not possible.
Got it. The definition quantity of the function is searched for.
Thanks Melody!
f(x)=1√x−99<x<∞
!
thank u for your help however it is x minus 9 not x to the negative ninth
How do you find the domain and range of 1/square root of x-9
I think that this is what you intended?
f(x)=1√x−9
Let see..
you cannot divide by 0 and the square root of a real number cannot be neg so the range must be greater than 0
x-9 cannot be negative so x>9
so the range is f(x)>0
and the domain is x>9
Here is the funtion