Here's the question you clicked on:
ikoreanx3
For the functions given, find a) (f*g)(x) b) (f*g)(2) f(x)=x-3, g(x)=x+6 a) (f*g)(x) = ? b) (f*g)(2)= ?
so, multiply them then
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?
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?
so it is written (x-3)*(x+6)(2)?
no not quite...where you see that (2) is exactly where you saw the (x) ....did you write the other one as (x-3)*(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"
So then... it would be (2)^2+3(2)-18 =