anonymous
  • anonymous
Let f(x) = \frac{1}{\sqrt{ x - 5 }} ,
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
input to squares must be non negative for real solution so x can not be less than. 5
anonymous
  • anonymous
do the work yourself lazy feather
anonymous
  • anonymous
b*tch

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anonymous
  • anonymous
\[f(x)={1\over\sqrt{x-5}}\]what about it? :)
austinL
  • austinL
\(\large{f(x) = \dfrac{1}{\sqrt{ x - 5 }}}\) That is a function, what are we supposed to do with this?
anonymous
  • anonymous
@brooklyn1, do you still need help with this?
anonymous
  • anonymous
I'm trying to find the domain and range of f(x)
anonymous
  • anonymous
I'm not sure where to begin
anonymous
  • anonymous
It says I have to find both domain and range algebraically
anonymous
  • anonymous
okay well when finding the domain, we can see that x cannot be less than or equal to 5, yes? because the radical is in the denominator, if x is 5, we will get a zero and that would make the function undefined. it can't be less than 5 because that would make the radical negative and that would be imaginary so we know domain begins after 5 \(\implies~domain=(5,\infty)\) now since the degree of the numerator is less than the denominator, we follow the asymptote rule and get that there is an asymptote at x=0 so there is a break in the range at 0 \(\implies~range=(-\infty,0)\cup(0,\infty)\)
anonymous
  • anonymous
Thank you so much yummydum
anonymous
  • anonymous
No problem!

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