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Mathematics 22 Online
OpenStudy (anonymous):

f(x)= sqrt (x+2) g(x)= sqrt (x-1) find (f*g)(x) (f/g)(x) also find the domain of each

OpenStudy (anonymous):

does the asterisk mean compostion?

OpenStudy (anonymous):

like the first part is f(g)

OpenStudy (anonymous):

you want \[f\circ g(x)\]?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

not sure how you did the little circle

OpenStudy (anonymous):

ok so it is \[f \circ g(x) = f(g(x))=f(\sqrt{x-1})\]

OpenStudy (anonymous):

=\[\sqrt{\sqrt{x-1}+2}\]

OpenStudy (anonymous):

yes then you have to substitute f(x) in for the x

OpenStudy (anonymous):

no you put g(x) where you see x in f(x)

OpenStudy (anonymous):

do it step by step as i did and you will not mess this up

OpenStudy (anonymous):

k i get the first part now :)

OpenStudy (anonymous):

what about f/g

OpenStudy (anonymous):

now you need domain yes?

OpenStudy (anonymous):

\[\frac{f}{g}\] is trivial, just put f on top of g

OpenStudy (anonymous):

k

OpenStudy (anonymous):

hey satellite, can u come to chat if u don't mind

OpenStudy (anonymous):

\[\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x+2}}{\sqrt{x-1}}\]

OpenStudy (anonymous):

now we do the domain. domain of f/g is x > 1 since the thing under the radical must be non - negative,and since it is in the denominator it cannot be 0 either. that is why you need the strict inequality.

OpenStudy (anonymous):

harder is first one

OpenStudy (anonymous):

@krish now that you mention it i do not see chat!

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