Question

how do i properly express the domain of the function f(x) = 2/sqrt(x^2-5) in set builder notation

Answer 1

\[f(x) = \frac{ 2 }{ \sqrt{x^{2}-5} }\]

Answer 2

i know the domain is all reals except \[\sqrt{5} and -\sqrt{5}\]

Answer 3

but how do I express it properly?

Answer 4

keep in mind that you cannot take the square root of a negative number

Answer 5

but wait if x is being squared, then it doesnt matter if x is negative, becuase -sqrt(5) squared is 5 right?

Answer 6

yes but x^2 - 5 is negative for some interval

Answer 7

you have to make sure x^2 - 5 is NOT zero and also NOT negative

Answer 8

the original question had x squared Plus 5, sorry

Answer 9

thats what i meant, i will redrwaw it

Answer 10

\[f(x) = \frac{ 2 }{ \sqrt{x^{2}+5} }\]

Answer 11

\[x = \left\{ x \in \mathbb{R} | \left| x \right| > \sqrt{5} \right\}\]

Answer 12

Oh, just seen what you said about it being +5, ignore the above.

Answer 13

you are correct though tom

Answer 14

that's the same as saying

x < -sqrt(5) or x > sqrt(5)

Answer 15

ok. Becuase in my test, i wrote the answer as the domain of \[D _{x}: \mathbb{R}/-\sqrt{5},\sqrt{5}\] and was marked wrong

Answer 16

i dont know why

Answer 17

yeah because you don't just kick out sqrt(5), -sqrt(5)

Answer 18

you have to kick out everything in between because those values make x^2 - 5 > 0 false

Answer 19

Thanks @jim_thompson5910, but surely the domain is all real numbers excluding +/-sqrt(5) now that it's x^2+5 in the denominator?

Answer 20

wait, suppose i was to kick out only the values of -sqrt(5) and sqrt(5) for some function g, how would i write that domain?

Answer 21

jim, i am no longer using the x^2 - 5, now its the x^2 + 5

Answer 22

oh it's been changed to +5, I see

Answer 23

\[\left\{ x \in \mathbb{R} | \left| x \right| \neq \sqrt{5}\right\}\]

Answer 24

so x^2 + 5 is ALWAYS positive

x^2 is nonnegative, adding a positive bumps it into an all positive region

Answer 25

x^2+5 being always positive means x^2 + 5 = 0 has no solutions

Answer 26

@Rizags, your thinking was correct but I think the problem is that you haven't written your answer in set builder notation. Maybe Jim can confirm this?

Answer 27

the domain is the set of all real numbers

there is no real number x that causes a division by zero error since x^2 + 5 = 0 has no real solutions

we also don't have to worry about taking the square root of a negative number

Answer 28

it isnt about set builder, the teacher said that my answer represented a disjunction of the values, but the answer should have been a conjuction, as in x cannot equal sqrt(5) AND -sqrt(5). Just not sure how to write that.

Answer 29

and she disregarded the fact that x^2 + 5 has no real solutions

Answer 30

Oops, I'm back on x^2-5 with my thinking, lol. Thanks for pointing that out, Jim.

Answer 31

the graph visually confirms it

Answer 32

Essentially, what I am asking is, for any function g, with domain of all values excluding sqrt(5) and -sqrt(5), how would i write the domain in set builder without using absolute value of x, as tom did above?

Answer 33

@Rizags I was wrong. What Jim is saying is that it's impossible to find a real number that doesn't work in your equation (try putting sqrt(5) in and you'll get sqrt(10)/5 as your output) - hence your domain is R, the real numbers.

Answer 34

And one more thing, is this the proper way to represent the limit of the function \[x ^{2} - 4x + 7\]
\[\lim_{h \rightarrow 0}\frac{ f(x+h) - f(x) }{ h } = 2x+4 \]

Answer 35

how would i express the difference quotient? is this coorect?

Answer 36

is that equation i wrote correct?

Answer 37

f(x) = x^2 - 4x + 7

you need to find f(x+h)

Answer 38

2x+4 is close, but incorrect

Answer 39

oh, its minus 4

Answer 40

yes

Answer 41

but is the equation format correct, because my teacher is very strict about that

Answer 42

would i write, when F(x) = blank, then the limit equation?

Answer 43

you have a true statement if you write

\[\Large \lim_{h \rightarrow 0}\frac{ f(x+h) - f(x) }{ h } = 2x-4\]

Answer 44

but, i have to identify f(x) first, right?

Answer 45

of course there are multiple steps in between, but it's still correct

Answer 46

f(x) = x^2 - 4x + 7

Answer 47

but do i identify that in my formal answer?

Answer 48

you find f(x+h) and then simplify and then find the limit

Answer 49

or is it implied?

Answer 50

I'm not sure what your teacher wants ultimately

Answer 51

but s/he probably wants the steps to get from A to Z

Answer 52

i know how to do it, its just i need a very clear representation of the answer. Would this be clear?

\[When f(x) = x^{2} - 4x + 7,\]
\[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h } = 2x - 4\]

Answer 53

yes that is clear and correct

Answer 54

I can't imagine she will have any problems with that, it looks good to me.

Answer 55

ok, thanks a lot for the help I appreciate it

Answer 56

np