Question

How do you find a domain?
for example: let f(x)=x+5 and g(x)=x^2-25. What is the domain of f/g (x)?

Answer 1

The function will be

x+5
----
x²-25

factoring the bottom, and canceling x+5,

f/g (x) = 1/(x-5)

Answer 2

The domain can be any thing, but the restriction is 5.

Answer 3

(-∞,5) Union (5,∞)

Answer 4

how would you factor the bottom

Answer 5

x²-25 = x² - 5² = (x-5)(x+5)

Answer 6

clear with that ?

Answer 7

I have finally started on limits officially, and this really reminds me one of those algebraically easy ones.

Answer 8

The domain should be found BEFORE simplifying f/g(x).
Therefore, domain is all real numbers except x = -5 and x = +5.

Answer 9

yes thank you

Answer 10

In case you need confirmation: http://www.wolframalpha.com/input/?i=domain+of+%28x%2B5%29%2F%28x^2-25%29

Answer 11

aum, I thought you cancel the x+5, don't you ?

Answer 12

You can simplify the expression. But when domain is asked for a rational expression you need to find what makes the denominator zero BEFORE simplifying.

Answer 13

well, all I can tell you....

\(\LARGE\color{black}{ \lim_{x \rightarrow C} ~\frac{x+5}{x^2-25} }\)

\(\LARGE\color{black}{ \lim_{x \rightarrow C} ~\frac{x+5}{(x+5)(x-5)} }\)

\(\LARGE\color{black}{ \lim_{x \rightarrow C} ~x+5 }\)

so limit x→5 and f(5), (I think) both exist)

Answer 14

wait, no they don't nvm

Answer 15

The limit exists, and the limit as x→C of f(x) = f(C)

Answer 16

and the sides are equal..

Answer 17

BUT f(C) does not exist.

Answer 18

The function is continuous at x = -5. No dispute there.
But when f(x) is divided by g(x) then all values that make g(x) zero have to be excluded from the domain.

Answer 19

\(\LARGE\color{black}{ \lim_{x \rightarrow -5} ~\frac{x+5}{x^2-25} }\)
I like more,

\(\LARGE\color{black}{ \lim_{x \rightarrow -5} ~\frac{\frac{d}{dx}(x+5)}{\frac{d}{dx}(x^2-25)} }\)

\(\LARGE\color{black}{ \lim_{x \rightarrow -5} ~\frac{1}{2x} }\)

\(\LARGE\color{black}{ \frac{1}{-10} }\)

Answer 20

continuous at x=... doesn't tend to make sense to me though, but I can somewhat see what you are saying.

Answer 21

L'H'S ?! just showing off.

Answer 22

Continuous function has to have 3 conditions.

f(c) exists
lim x→c f(c) exist
lim x→c f(x) = f(c)