f(x) = sqrt 5/x-2

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

f(x) = sqrt 5/x-2

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

- anonymous

Is it\[f(x)=\sqrt{\frac{5}{x}-2}\]or\[f(x)=\sqrt{{5}{x-2}}\]?

- anonymous

lol i wish i could write codes but naw :(

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

umm no let me draw it out

- anonymous

Blarg, second one was supposed to be\[f(x)=\sqrt{\frac{5}{x-2}}\]

- anonymous

yes that is correct

- anonymous

please go slowly :D

- anonymous

Ok. (I forgot to put the \frac in the first time.)
First, we'll start with assuming that the domain is all real numbers, and then restrict it as we go.
Now, we cannot have any values of x that would make 5/(x-2) be negative. The only way that could be negative though, would be if x was less than 2. So we know that x has to be greater than or equal to 2.
BUT when we look at the denominator (x-2) we know that we can't let it equal 0. So we can't let x equal 2, or we would end up dividing by 0.
So this leaves us with x is greater than 2. So the domain is\[2 < x < \infty\]
To get the range, we look at what happens to f(x) as we vary x throughout the domain. The closer and closer we get to 2, the smaller (x-2) becomes. And when we're dividing by an extremely small number we end up getting a very large number. Because there is no number that we can point to as being the next largest number from 2 (basically this means that if I were to try and claim that some number a was the next largest number from 2, you would always be able to point to a number that was between 2 and a). So as x goes to 2, f(x) goes to infinity.
A similar argument holds for when x goes to infinity. As we divide by ever larger numbers, we get very small numbers out. However, there's never going to be a way to divide 5 by a number and get exactly 0 out (we can only get close to it), so we can say that y will be greater than 0.
So the range is\[0 < y < \infty\]

- anonymous

hold on...getting there ..

- anonymous

so x cannot be negative or cannot be 0 ?

- anonymous

Well, anything inside of a square root sign cannot be negative.
Also anything on the bottom of a fraction can't be zero.
There are different restrictions for different functions.
\[f(x)=x^{2}\]has no such restrictions for example, while\[f(x)=\frac{5}{x^{2}-7}\]has the restriction that x^2-7 cannot be 0.

- anonymous

oh i see

- anonymous

so when it's under sqrt it's restricted

- anonymous

Right.

- anonymous

Ok, thanks for your detailed words :D
I must practice more to understand
Will get more helped if i need, thank you !!

- anonymous

You're welcome. I have to head back to my apartment now, but I'll be back online in about 20 minutes if you need more help then.

- anonymous

umm i think that's it for the night! i will do some science homework now lol !
thanks so much!!
sorry for wasted your time :(

- anonymous

You didn't waste my time. I enjoy helping people learn (that's why I'm here! :P)

Looking for something else?

Not the answer you are looking for? Search for more explanations.