Question

Let f(x) = \frac{1}{\sqrt{ x - 5 }} ,

Answer 1

input to squares must be non negative for real solution
so x can not be less than. 5

Answer 2

do the work yourself lazy feather

Answer 3

b*tch

Answer 4

\[f(x)={1\over\sqrt{x-5}}\]what about it? :)

Answer 5

\(\large{f(x) = \dfrac{1}{\sqrt{ x - 5 }}}\)
That is a function, what are we supposed to do with this?

Answer 6

@brooklyn1, do you still need help with this?

Answer 7

I'm trying to find the domain and range of f(x)

Answer 8

I'm not sure where to begin

Answer 9

It says I have to find both domain and range algebraically

Answer 10

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 \(\implies~range=(-\infty,0)\cup(0,\infty)\)

Answer 11

Thank you so much yummydum

Answer 12

No problem!