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ikoreanx3
Group Title
For the functions given, find a) (f*g)(x) b) (f*g)(2)
f(x)=x3, g(x)=x+6
a) (f*g)(x) = ?
b) (f*g)(2)= ?
 one year ago
 one year ago
ikoreanx3 Group Title
For the functions given, find a) (f*g)(x) b) (f*g)(2) f(x)=x3, g(x)=x+6 a) (f*g)(x) = ? b) (f*g)(2)= ?
 one year ago
 one year ago

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jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
so, multiply them then
 one year ago

johnanon93 Group TitleBest ResponseYou've already chosen the best response.1
so here, you literally just plug in those functions wherever you see f(x) and g(x) example (f*g)(x) you have an f(x), and you have a g(x)...plug them in (x  3) * (x + 6) can you do this?
 one year ago

ikoreanx3 Group TitleBest ResponseYou've already chosen the best response.0
x^2+3x18?
 one year ago

johnanon93 Group TitleBest ResponseYou've already chosen the best response.1
correct...that would be your a) now for b) you have (f*g)(2) ....notice how the (2) is where the (x) was before....so what do you think you do with that 2?
 one year ago

ikoreanx3 Group TitleBest ResponseYou've already chosen the best response.0
so it is written (x3)*(x+6)(2)?
 one year ago

johnanon93 Group TitleBest ResponseYou've already chosen the best response.1
no not quite...where you see that (2) is exactly where you saw the (x) ....did you write the other one as (x3)*(x+6)(x)? no because that wouldn't have worked. instead you used that (x) and you plugged it in for every x you saw in the equation a.k.a (x  3)*(x + 6) **same right? so now....you would plug (2) in for EVERY "x" you see in the equation you derived from finding (f*g)(x) remember how you had gotten x² + 3x  18 now...you would plug 2 into that equation wherever you see an "x"
 one year ago

ikoreanx3 Group TitleBest ResponseYou've already chosen the best response.0
So then... it would be (2)^2+3(2)18 =
 one year ago

johnanon93 Group TitleBest ResponseYou've already chosen the best response.1
correct! :)
 one year ago

ikoreanx3 Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much :)
 one year ago

johnanon93 Group TitleBest ResponseYou've already chosen the best response.1
no problem!
 one year ago
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