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# calculus

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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
+90212
+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
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#1
+6915
+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
+19
+1

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

medlockb1234  Sep 14, 2017
#3
+90212
+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

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