domain & endpoint for function

[snakeyesxlaw]

(f(x)) / (g(x))

= 1/5x

how do i find the domain & interval, endpoints for the function?

its known that the denominator can not equal 0, but how to find the endpoints or intervals?

 

[Deleted member 4993]

(f(x)) / (g(x))

= 1/5x

how do i find the domain & interval, endpoints for the function?

its known that the denominator can not equal 0, but how to find the endpoints or intervals?

What are the definitions of those terms?( domain & interval, endpoints )

Is your function
\(\displaystyle h(x) = \frac{1}{5}x\)

or

\(\displaystyle h(x) = \frac{1}{5x}\)

 

[snakeyesxlaw]

(f(x)) / (g(x))

= 1/5x

how do i find the domain & interval, endpoints for the function?

its known that the denominator can not equal 0, but how to find the endpoints or intervals?

What are the definitions of those terms?( domain & interval, endpoints )

Is your function
\(\displaystyle h(x) = \frac{1}{5}x\)

or

\(\displaystyle h(x) = \frac{1}{5x}\)

the function is:
\(\displaystyle h(x) = \frac{1}{5x}\)

domain is the x-ints
interval is what x can be
so i know x cant be 0, otherwise it would be undefined but it appears that it can be any other # but when submiting the answers as (-inf,0)U(0,inf), it is incorrect

 

[ilaggoodly]

oh, are you sure that the domain is the x-intercepts, sure its not "all possible x values for the equation" ?

 

[snakeyesxlaw]

oh, are you sure that the domain is the x-intercepts, sure its not "all possible x values for the equation" ?

thought it was opposite day for awhile, its actually all x values except x= values unless its [], but u didnt provide any suggestions toward the main part of the problem so ur response was essentially not useful at all..

 

[ilaggoodly]

sure i did, you would have gotten it wrong if ya thought the domain was the x intercepts what happened to self-discovery?

 

[snakeyesxlaw]

so my answer was, (-inf,0)U(0,inf)

and that was wrong.


here is the original values for f(x) & g(x):

Let f(x)= 1 / (x + 3) and g(x)= (5 x) / (x + 3)

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

stuck at the possibilities for the domain?

 

[Deleted member 4993]

so my answer was, (-inf,0)U(0,inf)

and that was wrong.


here is the original values for f(x) & g(x):

Let f(x)= 1 / (x + 3) and g(x)= (5 x) / (x + 3)

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

stuck at the possibilities for the domain?

In this case, x = -3 is not within the domain

 

[snakeyesxlaw]

Domain:
(_) ( -infinity , a ) union (b , infinity )
-> (_) ( -infinity , a ) union ( a , b ) union (b , infinity )
(_) ( -infinity , a ) union (a , infinity )
(_) ( -infinity , a ] union [ b , infinity )
(_) ( -infinity , infinity )


a = 0
b = -3

 

[pka]

Given that \(\displaystyle f(x) = \frac{1}{{x + 3}}\;\& \;g(x) = \frac{{5x}}{{x + 3}}\) then the domain of \(\displaystyle \frac{{f(x)}}{{g(x)}}\) is \(\displaystyle ( - \infty , - 3) \cup ( - 3,0) \cup (0,\infty )\).