Question

College algebra http://prntscr.com/mezyo9

Answer 1

For rational function domains, the denominator cannot equal 0

Answer 2

\(1-1x=0\)

Simplify (I assume you know how to do this)


\(x=1\)

The domain is all values except 1

\((-∞,1)\cup (1,∞) \)

Answer 3

isnt there supposed to be 3

Answer 4

hmm not sure how you got a 3

Answer 5

no like in a simlar problem the domain looked like (-infinity,0) U (0, 2/3) U (2/3, infinity)

Answer 6

Oh you mean like 3 different sets of intervals?


In this equation no, this only has 1 exception, if the equation were something like

\(\frac{1}{(x-6)(3x-2)}\), then it would have multiple

Answer 7

http://prntscr.com/mf04c8 How did they get this ?

Answer 8

Ah ok I see combined both domains from f(x) and g(x)

f(x) has a domain of (-∞,1)u(1,∞)
g(x) has a domain of (-∞,0)u(0,∞)

Technically \(f ~o~ g(x)\) is defined at 0
Which is why I did not include 0 in the domain section


Basically I messed up in that you have to include the domains of its components

Answer 9

I see I see, Im have a bit of trouble with part (b) now, do you think you could assist?

Answer 10

To do \(g~o~f\) we do the opposite by putting f into g(x)

\(f(x)=\boxed{\frac{1}{x-1}}\)
\(g(x)=\large \frac{1}{x}\)

\(\frac{1}{\boxed{\frac{1}{x-1}}}\) fraction in a fraction is just reciprocal


\(g~o~f(x)=x-1\)

Answer 11

Thank you appreciate it