do you have an idea for it yet?
(f + g)(x) that is short of adding f(x) + g(x)
so add 5x − 12 And 11x + 2
and f(x)+g(x) is short for: write down the "formula" for f(x) and add it to the "formula" for g(x) yes
yes i believe
B is the same idea except you multiply (use FOIL to multiply it out)
let me add the first one real quick hold on
can you do parts A and B?
Yea i think so, ill try could you check my work?
for part C f[g(x)] means "in the formula for f(x), replace "x" with g(x)"
for part B i got 55x2−122x−24 i think thats wrong
too bad, I got the same thing.
yes, you did it correctlly
ill just put that as the answer, because i honestly have to clue.
ooh yay okay
you can check that it is correct. pick a number for x, such as 1 then f(x) = 5x − 12 at x=1 is 5-12= -7 and g(x) = 11x + 2 at x=2 is 11+2 = 13 and -7*13= -91 now use your formula 55x^2−122x−24 at x=1... you get 55 -122 -24 = -91 same number. that does not prove it, but if you did that for a few values, and it always works, it probably is correct.
for part c, questions like that confuse me cause i never learned the foil method.
that one confuses me too
but if you FOIL it: ( 5x − 12) (11x+2) F first: 5x*11x = 55x^2 O outer: 5x*2 = 10x I inner: -12*11x = -132x L last: -12*2 = -24 combine the two middle terms 10x-132x = -122 x and the answer is 55x^2 -122x -24
part C f[g(x)] first write down f(x)
5x − 12
like this f(x)=5x − 12 we want f[g(x)] that means erase the x and replace it with g(x) erase and replace: f( g(x) ) = 5g(x) -12
then replace g(x) with g(x)'s formula (11x + 2) be sure to use the parens
can you replace g(x) in 5g(x) -12 with (11x+2)
f( 11x + 2) = 5(11x + 2) -12
yes, but we usually leave the left side in "short form" f(g(x)) and only expand the right side but yes that looks good. you can distribute the 5 (multiply 5 times each thing inside the parens) and simplify the right side a little.
the right side is 5(11x + 2) -12 do 5 * 11x + 5*2 to distribute the 5 55x + 10 -12 now simplify 10-12
yes. so part C is f[g(x)]= 55x-2