+0

# calculus

+1
38
3
+15

How do you find the domain and range of 1/square root of x-9

medlockb1234  Sep 14, 2017
edited by medlockb1234  Sep 14, 2017

#3
+90158
+2

How do you find the domain and range of 1/square root of x-9

I think that this is what you intended?

$$f(x)=\frac{1}{\sqrt{x-9}}$$

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

https://www.desmos.com/calculator/n3lq1imdho

Melody  Sep 14, 2017
Sort:

#1
+6899
+1

How do you find the range of 1/square root of x-9

Hello M!

$${\color{BrickRed}\frac{1}{\sqrt{x^{-9}}}}=\frac{1}{\sqrt{\frac{1}{x^9}}}=\sqrt{x^9}\color{blue}=x^4\sqrt{x}$$

sorry!
so

$${\color{BrickRed}\frac{1}{\sqrt{x-9}}}=\frac{\sqrt{x-9}}{x-9}$$

More is not possible.

Got it. The definition quantity of the function is searched for.

Thanks Melody!

$$f(x)=\frac{1}{\sqrt{x-9}}\\ \color{blue}9< x<\large ∞$$

!

asinus  Sep 14, 2017
edited by asinus  Sep 14, 2017
edited by asinus  Sep 14, 2017
#2
+15
+1

thank u for your help however it is x minus 9 not x to the negative ninth

medlockb1234  Sep 14, 2017
#3
+90158
+2

How do you find the domain and range of 1/square root of x-9

I think that this is what you intended?

$$f(x)=\frac{1}{\sqrt{x-9}}$$

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

https://www.desmos.com/calculator/n3lq1imdho

Melody  Sep 14, 2017

### 26 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details