Question

If f(x) = x^2 - 25 and g(x) = x - 5, what is the domain of (f/g)(x)?

Answer 1

\[f(x) = x^2-25\] and \[g(x) = x-5 \]
\[(\frac{f}{g})(x)\]
is this your question and are those the right functions?

Answer 2

Yes, that's correct.

Answer 3

alright so we need \[(\frac{f}{g})(x) \]
\[(\frac{f}{g})(x) = \frac{x^2-25}{x-5}\]
can you factor the \[x^2-25\]?
after that step we can cancel a term out

Answer 4

I don't know how to solve it, I'm not very good with math. lol.

Answer 5

oh ok. no problem . we can use the difference of squares formula
\[(a^2-b^2) = (a+b)(a-b) \]
so if we let a = x and b = 5
\[(x^2-5^2) = (x+5)(x-5) \]

Answer 6

so \[(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}\] so what does the numerator and denominator have in common

Answer 7

there's a x-5 in the numerator and denominator so that term is canceled out

Answer 8

\[(\frac{f}{g})(x) = x+5\]
since our new function isn't a fraction , there are no restrictions so the domain is all real numbers.

Answer 9

the graph should be a straight line and it's a function because it passes the vertical line test. A vertical line test is needed to determine if a graph is a function or not. If it's a function then the vertical line should hit the graph only once. If it's not a function, the vertical line crosses the graph more than once.

Answer 10

So my answer options are

all real values of x
all real values of x except x = 5
all real values of x except x = –5
all real values of x except x = 5 and x = –5

Would that mean it's a? Since they're all real numbers? Or would it be b?

Answer 11

has to be a... all reals. the function is not a fraction

Answer 12

Thank you! :)

Answer 13

this is the graph of the function (f/g) (x)

Answer 14

if you draw a vertical line on this graph, the vertical line only touches once, so it's a function .

Answer 15

hehe domain does not include x=5. It would not show up on a graph because its a hole.

If you graph \(\frac{x^2-25}{(x-5)}\) and zoomed in infinitely close it would look like this



\[\frac{x^2-25}{(x-5)}\ne x+5\] in general (one is defined for 5, one is not)

But if we remove \(5\) from the domain it is :)

Answer 16

if we factor the numerator, the x-5 cancels out leaving us with the new function no longer being a fraction. Hence, no restrictions... all reals.

\[(\frac{f}{g})(x) = \frac{x^2-25}{x-5} \]
\[(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}\]
\[(\frac{f}{g})(x) = x+5 \]

Answer 17

if you're not factoring then there is a restriction which is all reals except when x = 5. But I think that there have been times when canceling terms can happen.

Answer 18

both
\[(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5} \]
and
\[(\frac{f}{g})(x) = x+5 \]

produce different results..
It's like do we consider the factored version in that case it's all reals on the domain or the non-factored case all reals except x = 5 ?